An angle is considered as the figure obtained by rotating a given ray about its end-point. Two angles are said to be complementary angles if their sum is equal to \({90^ \circ }\). In this article, we will learn about Trigonometric Ratios of Complementary Angles and how to find them.

Trigonometry is a branch of mathematics that deals with the measurement of angles and the problems based on allied angles. It implies the study of relationships between the sides and angles of a triangle. Read on to find more about Trigonometry and Trigonometric Ratios.

Trigonometry

Just imagine if you are on top of a building, and at a certain distance from the foot of the building, you find your car parked. Can you tell the distance between the foot of a tower and the car parked? A right triangle is imagined to be made in this situation. In such a situation, the distances or heights can be found using mathematical techniques, which come under a branch of mathematics called Trigonometry.

The word trigonometry is derived from the Greek word tri (meaning three), gon (meaning sides) and metron (meaning measure). Therefore, trigonometry can be defined as the study of relationships between the sides and angles of a triangle.

Early astronomers used it to find out the distances of the stars and planets from the Earth. Most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.

Trigonometric Ratios

As discussed, trigonometry is a concept to find the remaining sides and angles of a triangle when some sides and angles are given. This problem is solved by using some ratios of the sides of a triangle concerning its acute angles. These ratios of acute angles are called trigonometric ratios of angles. Let us now define various trigonometric ratios.

Let us take a right-angled triangle \(\Delta ABC\) as shown.

Here, \(∠CAB\) or \(∠A\) is an acute angle. The side \(BC\) is facing \(∠A\). So, we call it the side opposite to \(∠A. AC\) is the hypotenuse of the triangle \(ABC\), and the side \(AB\) is a part of \(∠A.\) So, we call it the side adjacent to \(∠A\).

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The trigonometric ratios of the angle A in the right triangle \(ABC\) are defined as follows:

Sine of \(\angle A = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}}{{{\rm{hypotenuse}}}} = \frac{{BC}}{{AC}}\)

Cosine of \(\angle A = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}}{{{\rm{hypotenuse}}}} = \frac{{AB}}{{AC}}\)

Tangent of \(\angle A = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}} = \frac{{BC}}{{AB}}\)

Cosecant of \(\angle A = \frac{1}{{{\rm{sine}}\,{\rm{of}}\,\angle A}}\frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}} = \frac{{AC}}{{BC}}\)

Secant of \(\angle A = \frac{1}{{{\rm{cosine}}\,{\rm{of}}\,\angle A}}\frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}} = \frac{{AC}}{{AB}}\)

Cotangent of \(\angle A = \frac{1}{{{\rm{tangent}}\,{\rm{of}}\,\angle A}}\frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}} = \frac{{AB}}{{AC}}\)

The ratios defined above are abbreviated as \({\rm{sinA, cosA, tanA, cosecA, secA}}\) and \(\cot \,A,\) respectively. The ratios \({\rm{cosecA, secA}}\) and (\cot \,A,\) are respectively the reciprocals of the ratios \(\sin \,A,\,\cos \,A\) and \(\tan \,A.\)

Also, observe that \(\tan \,A = \frac{{BC}}{{AB}} = \frac{{\frac{{BC}}{{AC}}}}{{\frac{{AB}}{{AC}}}} = \frac{{\sin \,A}}{{\cos \,A}}\) and \(\cot \,A = \frac{{\cos \,A}}{{\sin \,A}}\)

So, the trigonometric ratios of an acute angle in a right-angled triangle express the relationship between the angle and length of its sides.

Relationship Between Trigonometric Ratios

Below is the list of the relationship between some of the trigonometric ratios.

1. \(\tan A = \frac{{\sin A}}{{\cos A}}\)

2. \(\cot A = \frac{{\cos A}}{{\sin A}} \Rightarrow \cot A = \frac{1}{{\tan A}}\)

3. \({\mathop{\rm cosec}\nolimits} A = \frac{1}{{{\mathop{\rm sinA}\nolimits} }} \Rightarrow \sin A = \frac{1}{{{\mathop{\rm cosec}\nolimits} A}}\)

4. \(\sec A = \frac{1}{{\cos A}} \Rightarrow \cos A = \frac{1}{{\sec A}}\)

Identities Between Trigonometric Ratios

An equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle(s) involved.

Below are some of the identities between the trigonometric ratios.

1. \({\sin ^2}A + {\cos ^2}A = 1\)

2. \({\sec ^2}A – {\tan ^2}A = 1\)

3. \({{\mathop{\rm cosec}\nolimits} ^2}A = 1 + {\cot ^2}A\)

Trigonometric Ratios of Some Specific Angles

In this section, we will know the values of the trigonometric ratios of the angles \({0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }\) and \({90^ \circ }\) which are derived using Pythagoras theorem in a right-angled triangle.

Below is the table of all the values of trigonometric ratios of \({0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }\) and \({90^ \circ }\).

\(∠A\)	\({0^ \circ }\)	\({30^ \circ }\)	\({45^ \circ }\)	\({60^ \circ }\)	\({90^ \circ }\)
\({\rm{SinA}}\)	\(0\)	\(\frac{1}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{{\sqrt 3 }}{2}\)	\(1\)
\({\rm{CosA}}\)	\(1\)	\(\frac{{\sqrt 3 }}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{1}{2}\)	\(0\)
\({\rm{tanA}}\)	\(0\)	\(\frac{1}{{\sqrt 3 }}\)	\(1\)	\(\sqrt 3 \)	Not defined
\({\rm{CosecA}}\)	Not defined	\(2\)	\(\sqrt 2 \)	\(\frac{2}{{\sqrt 3 }}\)	\(1\)
\({\rm{SecA}}\)	\(1\)	\(\frac{2}{{\sqrt 3 }}\)	\(\sqrt 2 \)	\(2\)	Not defined
\({\rm{CotA}}\)	Not defined	\(\sqrt 3\)	\(1\)	\(\frac{1}{{\sqrt 3 }}\)	\(0\)

Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum is \({90^ \circ }\). 

In the \(\Delta ABC,\angle A + \angle C = {90^ \circ }.\)

\(\sin A = \frac{{BC}}{{AC}},\cos A = \frac{{AB}}{{AC}},\tan A = \frac{{BC}}{{AB}},{\mathop{\rm cosec}\nolimits} A = \frac{{AC}}{{BC}},\sec A = \frac{{AC}}{{AB}},\cot A = \frac{{AB}}{{BC}}\)

Now, let us write trigonometric ratios for \(\angle C = {90^ \circ } – \angle A\)

The opposite side of \({90^ \circ } – A\) is \(AB\), and the adjacent side of \({90^ \circ } – A\) is \(BC\).

Therefore, 

\(\sin \left( {{{90}^ \circ } – A} \right) = \frac{{AB}}{{AC}},\cos \left( {{{90}^ \circ } – A} \right) = \frac{{BC}}{{AC}},\tan \left( {{{90}^ \circ } – A} \right) = \frac{{AB}}{{BC}},{\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – A} \right) = \frac{{AC}}{{AB}},\sec \left( {{{90}^ \circ } – A} \right) = \frac{{AC}}{{BC}},\cot \left( {{{90}^ \circ } – A} \right) = \frac{{BC}}{{AB}} \ldots \ldots .({\rm{ii}})\)

Comparing the ratios in (i) and (ii), we get

\(\sin \left( {{{90}^ \circ } – A} \right) = \frac{{AB}}{{AC}} = \cos A\) and \(\cos \left( {{{90}^{\rm{o}}} – A} \right) = \frac{{BC}}{{AC}} = \sin \,A\)

Also, \(\tan \left( {{{90}^ \circ } – A} \right) = \frac{{AB}}{{BC}} = \cot A\) and \(\cot \left( {{{90}^{\rm{o}}} – A} \right) = \frac{{BC}}{{AB}} = \tan \,A\)

So, \(\sin \left( {{{90}^ \circ } – A} \right) = \cos A,\cos \left( {{{90}^ \circ } – A} \right) = \sin A\)

\(\tan \left( {{{90}^ \circ } – A} \right) = \cot A,\cot \left( {{{90}^ \circ } – A} \right) = \tan A\)

\(\sec \left( {{{90}^ \circ } – A} \right) = {\mathop{\rm cosec}\nolimits} A,{\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – A} \right) = \sec A\)

Solved Examples – Trigonometric Ratios of Complementary Angles

Q.1. Evaluate the following:
a. \(\frac{{\cos {{37}^ \circ }}}{{\sin {{53}^ \circ }}}\)
b. \(\frac{{\tan {{68}^ \circ }}}{{\cot {{22}^ \circ }}}\)
c. \(\frac{{\sin {{41}^ \circ }}}{{\cos {{49}^ \circ }}}\)
d. \(\frac{{{\mathop{\rm cosec}\nolimits} {{32}^ \circ }}}{{\sec {{58}^ \circ }}}\)

Ans: a. Given: \(\frac{{\cos {{37}^ \circ }}}{{\sin {{53}^ \circ }}}\)
We know that, \(\cos \left( {{{90}^ \circ } – A} \right) = \sin A\)
\( \Rightarrow \frac{{\cos {{37}^ \circ }}}{{\sin {{53}^ \circ }}} = \frac{{\cos \left( {{{90}^ \circ } – {{53}^ \circ }} \right)}}{{\sin {{53}^ \circ }}}\)
\( \Rightarrow \frac{{\sin {{53}^ \circ }}}{{\sin {{53}^ \circ }}} = 1\)
Therefore, \(\frac{{\cos {{37}^ \circ }}}{{\sin {{53}^ \circ }}} = 1\)

b. Given: \(\frac{{\tan {{68}^ \circ }}}{{\cot {{22}^ \circ }}}\)
We know that \(\tan \left( {{{90}^ \circ } – A} \right) = \cot A\)
\( \Rightarrow \frac{{\tan {{68}^ \circ }}}{{\cot {{22}^ \circ }}} = \frac{{\tan \left( {{{90}^ \circ } – {{22}^ \circ }} \right)}}{{\cot {{22}^ \circ }}}\)
\( = \frac{{\cot {{22}^ \circ }}}{{\cot {{22}^ \circ }}} = 1\)
Therefore, \(\frac{{\tan {{68}^ \circ }}}{{\cot {{22}^ \circ }}} = 1\)

c. Given: \(\frac{{\sin {{41}^ \circ }}}{{\cos {{49}^ \circ }}}\)
We Know that \(\sin \left( {{{90}^ \circ } – A} \right) = \cos A\)
\( \Rightarrow \frac{{\sin {{41}^ \circ }}}{{\cos {{49}^ \circ }}} = \frac{{\sin \left( {{{90}^ \circ } – {{49}^ \circ }} \right)}}{{\cos {{49}^ \circ }}}\)
\( = \frac{{\cos {{49}^ \circ }}}{{\cos {{49}^ \circ }}} = 1\)
Therefore, \(\frac{{\sin {{41}^ \circ }}}{{\cos {{49}^ \circ }}} = 1\)

d. Given: \(\frac{{{\mathop{\rm cosec}\nolimits} {{32}^ \circ }}}{{\sec {{58}^ \circ }}}\)
We know that \({\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – A} \right) = \sec A\)
\(\Rightarrow \frac{{{\mathop{\rm cosec}\nolimits} {{32}^ \circ }}}{{\sec {{58}^ \circ }}} = \frac{{{\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – {{58}^ \circ }} \right)}}{{\sec {{58}^ \circ }}}\)
\( = \frac{{\sec {{58}^ \circ }}}{{\sec {{58}^ \circ }}} = 1\)
Therefore, \(\frac{{{\mathop{\rm cosec}\nolimits} {{32}^ \circ }}}{{\sec {{58}^ \circ }}} = 1\)

Q.2. Evaluate the following:
a. \(\cos {48^ \circ } – \sin {42^ \circ }\)
b. \({\mathop{\rm cosec}\nolimits} {25^ \circ } – \sec {65^ \circ }\)
c. \(\frac{{\sin {{36}^ \circ }}}{{\cos {{54}^ \circ }}} – \frac{{\sin {{54}^ \circ }}}{{\cos {{36}^ \circ }}}\)
d. \({\cos ^2}{13^ \circ } – {\sin ^2}{77^ \circ }\)

Ans: a. Given, \(\cos {48^ \circ } – \sin {42^ \circ }\)
We know that \(\sin \left( {{{90}^ \circ } – A} \right) = \cos A\)
\( \Rightarrow \cos {48^ \circ } – \sin {42^ \circ } = \cos {48^ \circ } – \sin \left( {{{90}^ \circ } – {{48}^ \circ }} \right)\)
\( = \cos {48^ \circ } – \cos {48^ \circ }\)
\(=0\)
Therefore, \(\cos {48^ \circ } – \sin {42^ \circ } = 0\)

b. \({\mathop{\rm cosec}\nolimits} {25^ \circ } – \sec {65^ \circ }\)
We Know that \(\sec \left( {{{90}^ \circ } – A} \right) = {\mathop{\rm cosec}\nolimits} A\)
\( \Rightarrow {\mathop{\rm cosec}\nolimits} {25^ \circ } – \sec {65^ \circ } = {\mathop{\rm cosec}\nolimits} {25^ \circ } – \sec \left( {{{90}^ \circ } – {{25}^ \circ }} \right)\)
\( = {\mathop{\rm cosec}\nolimits} {25^ \circ } – {\mathop{\rm cosec}\nolimits} {25^ \circ }\)
\(=0\)
Therefore, \({\mathop{\rm cosec}\nolimits} {25^ \circ } – \sec {65^ \circ } = 0\)

c. \(\frac{{\sin {{36}^ \circ }}}{{\cos {{54}^ \circ }}} – \frac{{\sin {{54}^ \circ }}}{{\cos {{36}^ \circ }}}\)
We know that, \(\cos \left( {{{90}^ \circ } – A} \right) = \sin A\)
\( \Rightarrow \frac{{\sin {{36}^ \circ }}}{{\cos {{54}^ \circ }}} – \frac{{\sin {{54}^ \circ }}}{{{\mathop{\rm cos}\nolimits} {{36}^ \circ }}}\)
\( = \frac{{\sin {{36}^ \circ }}}{{\cos \left( {{{90}^ \circ } – {{36}^ \circ }} \right)}} – \frac{{\sin {{54}^ \circ }}}{{\cos \left( {{{90}^ \circ } – {{54}^ \circ }} \right)}}\)
\( = \frac{{\sin {{36}^ \circ }}}{{\sin {{36}^ \circ }}} – \frac{{\sin {{54}^ \circ }}}{{\sin {{54}^ \circ }}}\)
\(=1-1\)
\(=0\)
Therefore, \(\frac{{\sin {{36}^ \circ }}}{{\cos {{54}^ \circ }}} – \frac{{\sin {{54}^ \circ }}}{{\cos {{36}^ \circ }}} = 0\)

d. \({\cos ^2}{13^ \circ } – {\sin ^2}{77^ \circ }\)
We know that, \(\cos \left( {{{90}^ \circ } – A} \right) = \sin A\)
Also. \({\cos ^2}{13^{\rm{o}}} = {\left( {\cos \,{{13}^{\rm{o}}}} \right)^2}\) and \({\sin ^2}{77^{\rm{o}}} = {\left( {\sin \,{{77}^{\rm{o}}}} \right)^2}\)
So, \(\cos {13^ \circ } = {\left[ {\cos \left( {{{90}^ \circ } – {{77}^ \circ }} \right)} \right]^2} = {\left( {\sin {{77}^ \circ }} \right)^2}\)
\( \Rightarrow {\cos ^2}{13^ \circ } – {\sin ^2}{77^ \circ } = {\sin ^2}{77^ \circ } – {\sin ^2}{77^ \circ } = 0\)
Therefore, \({\cos ^2}{13^ \circ } – {\sin ^2}{77^ \circ } = 0\)

Q.3. If \(\sin 3A = \cos \left( {A – {{26}^ \circ }} \right)\), where \(3A\) is an acute angle, find the value of A.

Ans:  We are given that \(\sin 3A = \cos \left( {A – {{26}^ \circ }} \right)\)
We know that, \(\sin 3A = \cos \left( {{{90}^ \circ } – 3A} \right)\)
So, \(\sin 3A = \cos \cos \left( {A – {{26}^ \circ }} \right)\)
\( \Rightarrow \cos \left( {{{90}^ \circ } – 3A} \right) = \cos \left( {A – {{26}^ \circ }} \right)\)

Since \(\left( {{{90}^ \circ } – 3A} \right)\) and \({A – {{26}^ \circ }}\) are both acute angles, therefore, 
\(\left( {{{90}^ \circ } – 3A} \right) = \left( {A – {{26}^ \circ }} \right)\)
\( \Rightarrow 90 + 26 = 4A\)
\( \Rightarrow 4A = 116\)
\( \Rightarrow A = {29^ \circ }\)
Therefore, the value of A is \({29^ \circ }.\)

Q.4. Prove the following:
a. \(\sin {43^ \circ }\cos {47^ \circ } + \cos {43^ \circ }\sin {47^ \circ } = 1\)
b. \(\cos {38^ \circ }\cos {52^ \circ } – \sin {38^ \circ }\sin {52^ \circ } = 0\)
c. \(\sec {50^ \circ }\sin {40^ \circ } + \cos {40^ \circ }{\mathop{\rm cosec}\nolimits} {50^ \circ } = 2\)
d. \(\sec {70^ \circ }\sin {20^ \circ } – \cos {20^ \circ }{\mathop{\rm cosec}\nolimits} {70^ \circ } = 0\)

Ans: a. To Prove: \(\sin {43^ \circ }\cos {47^ \circ } + \cos {43^ \circ }\sin {47^ \circ } = 1\)
We Know that \(\sin \left( {{{90}^ \circ } – A} \right) = \cos A\) and \(cos \left( {{{90}^ \circ } – A} \right) = \sin A\)
So, \(\sin {43^ \circ }\cos {47^ \circ } + \cos {43^ \circ }\sin {47^ \circ } = \sin \left( {{{90}^ \circ } – {{47}^ \circ }} \right)\cos {47^ \circ } + \cos \left( {{{90}^ \circ } – {{47}^ \circ }} \right)\sin {47^ \circ }\)
\(= \cos {47^ \circ } \times \cos {47^ \circ } + \sin {47^ \circ } \times \sin {47^ \circ }\)
\(= {\cos ^2}{47^ \circ } + {\sin ^2}{47^ \circ }\)
\( = 1\left[ {{{\sin }^2}A + {{\cos }^2}A = 1} \right]\)
So. \(\sin {43^ \circ }\cos {47^ \circ } + \cos {43^ \circ }\sin {47^ \circ } = 1\)

b. To prove \(\cos {38^ \circ }\cos {52^ \circ } – \sin {38^ \circ }\sin {52^ \circ } = 0\)
We know that \(\sin \left( {{{90}^ \circ } – A} \right) = \cos A\) and \(cos \left( {{{90}^ \circ } – A} \right) = \sin A\)
So, \(\cos {38^ \circ }\cos {52^ \circ } – \sin {38^ \circ }\sin {52^ \circ } = \cos \left( {{{90}^ \circ } – {{52}^ \circ }} \right)\cos {52^ \circ } – \sin \left( {{{90}^ \circ } – {{52}^ \circ }} \right)\sin {52^ \circ }\)
\( = \sin {52^ \circ }\cos {52^ \circ } – \cos {52^ \circ }\sin {52^ \circ }\)
\(=0\)
So, \(\cos {38^ \circ }\cos {52^ \circ } – \sin {38^ \circ }\sin {52^ \circ } = 0\)

c. To prove: \(\sec {50^ \circ }\sin {40^ \circ } + \cos {40^ \circ }{\mathop{\rm cosec}\nolimits} {50^ \circ } = 2\)
We know that \(\sec \left( {{{90}^{\rm{o}}} – A} \right) = {\rm{cosec}}\,A\) and \({\rm{cosec}}\,\left( {{{90}^{\rm{o}}} – A} \right) = \sec \,A\)
So. \(\sec {50^ \circ }\sin {40^ \circ } + \cos {40^ \circ }{\mathop{\rm cosec}\nolimits} {50^ \circ } = \sec \left( {{{90}^ \circ } – {{40}^ \circ }} \right)\sin {40^ \circ } + \cos {40^ \circ }{\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – {{40}^ \circ }} \right)\)
\( = {\mathop{\rm cosec}\nolimits} {40^ \circ }\sin {40^ \circ } + \cos {40^ \circ }\sec {40^ \circ }\)
Also, \({\rm{cosec}}\,A = \frac{1}{{\sin \,A}}\) and \(\sec \,A = \frac{1}{{\cos \,A}}\)
\( \Rightarrow {\mathop{\rm cosec}\nolimits} {40^ \circ }\sin {40^ \circ } + \cos {40^ \circ }\sec {40^ \circ } = \frac{1}{{\sin {{40}^ \circ }}} \times \sin {40^ \circ } + \cos {40^ \circ } \times \frac{1}{{\cos {{40}^ \circ }}}\)
\(=1+1\)
\(=2\)
Therefore, \(\sec {50^ \circ }\sin {40^ \circ } + \cos {40^ \circ }{\mathop{\rm cosec}\nolimits} {50^ \circ } = 2\)

d. To prove: \(\sec {70^ \circ }\sin {20^ \circ } – \cos {20^ \circ }{\mathop{\rm cosec}\nolimits} {70^ \circ } = 0\)
We Know that \(\sec \left( {{{90}^{\rm{o}}} – A} \right) = {\rm{cosec}}\,A\) and \({\rm{cosec}}\left( {{{90}^{\rm{o}}} – A} \right) = \sec \,A\)
So, \(\sec {70^ \circ }\sin {20^ \circ } – \cos {20^ \circ }{\mathop{\rm cosec}\nolimits} {70^ \circ } = \sec \left( {{{90}^ \circ } – {{20}^ \circ }} \right)\sin {20^ \circ } – \cos {20^ \circ }{\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – {{20}^ \circ }} \right)\)
\( = {\mathop{\rm cosec}\nolimits} {20^ \circ }\sin {20^ \circ } – \cos {20^ \circ }\sec {20^ \circ }\)
Also, \({\rm{cosec}}\,A = \frac{1}{{\sin \,A}}\) and \(\sec \,A = \frac{1}{{\cos \,A}}\)
\( \Rightarrow {\mathop{\rm cosec}\nolimits} {20^ \circ }\sin {20^ \circ } – \cos {20^ \circ }\sec {20^ \circ } = \frac{1}{{\sin {{20}^ \circ }}} \times \sin {20^ \circ } – \cos {20^ \circ } \times \frac{1}{{\cos {{20}^ \circ }}}\)
\(=1-1\)
\(=0\)
So, \(\sec {70^ \circ }\sin {20^ \circ } – \cos {20^ \circ }{\mathop{\rm cosec}\nolimits} {70^ \circ } = 0\)

Q.5. Prove that \(\frac{{\cos {{70}^ \circ }}}{{\sin {{20}^ \circ }}} + \frac{{\cos {{59}^ \circ }}}{{\sin {{31}^ \circ }}} – 8{\sin ^2}{30^ \circ } = 0\)
Ans: To prove: \(\frac{{\cos {{70}^ \circ }}}{{\sin {{20}^ \circ }}} + \frac{{\cos {{59}^ \circ }}}{{\sin {{31}^ \circ }}} – 8{\sin ^2}{30^ \circ } = 0\)
LHS. \(\frac{{\cos {{70}^ \circ }}}{{\sin {{20}^ \circ }}} + \frac{{\cos {{50}^ \circ }}}{{\sin {{31}^ \circ }}} – 8{\sin ^2}{30^ \circ }\)

We Know that \(\cos \left( {{{90}^ \circ } – A} \right) = \sin A\)
So, \(\frac{{\cos {{70}^ \circ }}}{{\sin {{20}^ \circ }}} + \frac{{\cos {{59}^ \circ }}}{{\sin {{31}^ \circ }}} – 8{\sin ^2}{30^ \circ } = \frac{{\cos \left( {{{90}^ \circ } – {{20}^ \circ }} \right)}}{{\sin {{20}^ \circ }}} + \frac{{\cos \left( {{{90}^ \circ } – {{31}^ \circ }} \right)}}{{\sin {{31}^ \circ }}} – 8{\sin ^2}{30^ \circ }\)
\( = \frac{{\sin {{20}^ \circ }}}{{\sin {{20}^ \circ }}} + \frac{{\sin {{31}^ \circ }}}{{\sin {{31}^ \circ }}} – 8{\sin ^2}{30^ \circ }\)

We Know that, \(\sin {30^ \circ } = \frac{1}{2}\)
\( \Rightarrow 1 + 1 – 8{\left( {\frac{1}{2}} \right)^2}\)
\( = 2 – 8\left( {\frac{1}{4}} \right)\)
\(=2-2\)
\(=0=RHS\)
Therefore, \(\frac{{\cos {{70}^ \circ }}}{{\sin {{20}^ \circ }}} + \frac{{\cos {{59}^ \circ }}}{{\sin {{31}^ \circ }}} – 8{\sin ^2}{30^ \circ } = 0\)

Q.6. If \(\tan A = \cot B,\), Prove that \(A + B = {90^ \circ }\)
Ans: Given \(\tan A = \cot B\)
We Know that, \(\cot B = \tan \left( {{{90}^ \circ } – B} \right)\)
\( \Rightarrow \tan A = \tan \left( {{{90}^ \circ } – B} \right)\)
Since, A and B are acute angles,
\(A = {90^ \circ } – B\)
\( \Rightarrow A + B = {90^ \circ }\)
Hence, proved

Q.7.  If \(A, B\) and \(C\) are interior angles of a triangle \(ABC\), then show that \(\sin \left( {\frac{{B + C}}{2}} \right) = \cos \frac{A}{2}\)
Ans: We know that sum of interior angles of a triangle is \({180^ \circ }\)
Hence, \(\angle A + \angle B + \angle C = {180^{\rm{o}}}\) or \(A + B + C = {180^{\rm{o}}}\)
\( \Rightarrow B + C = {180^ \circ } – A\)
Multiply both sides by \(\frac{1}{2}\) we get
\(\frac{1}{2}(B + C) = \frac{1}{2}\left( {{{180}^ \circ } – A} \right)\)
\( \Rightarrow \frac{1}{2}(B + C) = {90^ \circ } – \frac{A}{2}\)
\( \Rightarrow \sin \left( {\frac{{B + C}}{2}} \right) = \sin \left( {{{90}^ \circ } – \frac{A}{2}} \right)\)
We know that \(\sin \left( {{{90}^ \circ } – A} \right) = \cos A\)
So, \(\sin \left( {\frac{{B + C}}{2}} \right) = \cos \frac{A}{2}\)

Q.8. If \(x\tan {15^ \circ }\tan {45^ \circ }\tan {60^ \circ }\tan {75^ \circ } = 1\). Find the value of \(x\)
Ans: Given \(x\tan {15^ \circ }\tan {45^ \circ }\tan {60^ \circ }\tan {75^ \circ } = 1\)
We Know that \(\tan {45^ \circ } = 1\)
So, \(x\tan {15^ \circ }\tan {60^ \circ }\tan {75^ \circ } = 1\)
\( \Rightarrow x \times \tan {15^ \circ } \times \tan \left( {{{90}^ \circ } – {{30}^ \circ }} \right) \times \tan \left( {{{90}^ \circ } – {{15}^ \circ }} \right) = 1\)
\( \Rightarrow x \times \tan {15^ \circ } \times \cot {30^ \circ } \times \cot {15^ \circ } = 1\)
\( \Rightarrow x \times \tan {15^ \circ } \times \frac{1}{{\tan {{30}^ \circ }}} \times \frac{1}{{\tan {{15}^ \circ }}} = 1\)
\( \Rightarrow x = \tan {30^ \circ }\)
Therefore, the value of \( \Rightarrow x = \tan {30^ \circ }\)

Summary

In the above article, we have learned the meaning of trigonometry, uses, and trigonometric ratios of different angles. We have also known the relationship between trigonometric ratios and the identities between these ratios.

Also, we have learned to find the trigonometric ratios of complementary angles and solved some example problems based on the trigonometric ratios of complementary angles.

FAQ – Trigonometric Ratios of Complementary Angles

Q.1. What are trigonometric ratios of complementary angles?
Ans: Two angles are said to be complementary if their sum equals \({90^ \circ }\). The relationship between the acute angle and the lengths of sides of a right-angled triangle is expressed by the trigonometric ratios. So, the trigonometric ratio of complementary angle is a trigonometric ratio of an angle obtained by subtracting it from \({90^ \circ }\).
Example: \(\sin \left( {{{90}^ \circ } – \theta } \right) = \cos \theta \). Here \(\sin \left( {{{90}^ \circ } – \theta } \right)\) is the trigonometric ratio of a complementary angle.

Q.2. How do you find the ratio of a complementary angle?
Ans: Imagine a right-angled triangle \(ABC\), right-angled at \(B\). So,\(\angle B = {90^ \circ }\) and \(\angle A + \angle C = {90^ \circ } \Rightarrow \angle C = {90^ \circ } – \angle A\)
We know that, in a right-angled triangle 
Sine of \(\angle A = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}}{{{\rm{hypotenuse}}}} = \frac{{BC}}{{AC}}\)
Cosine of \(\angle A = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}}{{{\rm{hypotenuse}}}} = \frac{{AB}}{{AC}}\)
Tangent of \(\angle A = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}} = \frac{{BC}}{{AB}}\)
Cosecant of \(\angle A = \frac{1}{{{\rm{sine}}\,{\rm{of}}\,\angle A}}\frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}} = \frac{{AC}}{{BC}}\)
Secant of \(\angle A = \frac{1}{{{\rm{cosine}}\,{\rm{of}}\,\angle A}}\frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}} = \frac{{AC}}{{AB}}\)
Cotangent of \(\angle A = \frac{1}{{{\rm{tangent}}\,{\rm{of}}\,\angle A}}\frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{\rm{A}}}} = \frac{{AB}}{{AC}}\)
Now, if we find the ratios of \(\angle C = {90^ \circ } – \angle A\) We get
\(\sin \left( {{{90}^ \circ } – A} \right) = \frac{{AB}}{{AC}},\cos \left( {{{90}^ \circ } – A} \right) = \frac{{BC}}{{AC}},\tan \left( {{{90}^ \circ } – A} \right) = \frac{{AB}}{{BC}},{\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – A} \right) = \frac{{AC}}{{AB}},\sec \left( {{{90}^ \circ } – A} \right) = \frac{{AC}}{{BC}},\cot \left( {{{90}^ \circ } – A} \right) = \frac{{BC}}{{AB}}\)
So, on comparing the above ratios, we get
\(\sin \left( {{{90}^ \circ } – A} \right) = \cos A,\cos \left( {{{90}^ \circ } – A} \right) = {\mathop{\rm sinA}\nolimits}, \tan \left( {{{90}^ \circ } – A} \right) = \cot A,\cot \left( {{{90}^ \circ } – A} \right) = \tan A\)
\(\sec \left( {{{90}^ \circ } – A} \right) = {\mathop{\rm cosecA}\nolimits} ,{\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – A} \right) = \sec A\)

Q.3. What trigonometric ratios are complementary?
Ans: We can write all six trigonometric ratios as complementary.
\(\sin \left( {{{90}^ \circ } – A} \right) = \cos A,\cos \left( {{{90}^ \circ } – A} \right) = \sin A\)
\(\tan \left( {{{90}^ \circ } – A} \right) = \cot A,\cot \left( {{{90}^ \circ } – A} \right) = \tan A\)
\(\sec \left( {{{90}^ \circ } – A} \right) = {\mathop{\rm cosec}\nolimits} A,{\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – A} \right) = \sec A\)

Q.4. What is the complement of \(\tan {30^ \circ }\)?
Ans: We Know that \(\cot \left( {{{90}^ \circ } – A} \right) = \tan A\)
So, \(\cot \left( {{{90}^ \circ } – {{30}^ \circ }} \right) = \tan {30^ \circ }\)
\( \Rightarrow \cot {60^ \circ } = \tan {30^ \circ }\)

Q.5. What are the examples of trigonometric ratios of complementary angles?
Ans: Below is some of the examples of trigonometric ratios of complementary angles.
\(\sin {18^ \circ } = \sin \left( {{{90}^ \circ } – {{72}^ \circ }} \right) = \cos {72^ \circ }\)
\(\tan {26^ \circ } = \tan \left( {{{90}^ \circ } – {{64}^ \circ }} \right) = \cot {64^ \circ }\)
\({\mathop{\rm cosec}\nolimits} {31^ \circ } = {\mathop{\rm cosec}\nolimits} \left( {{{90}^ \circ } – {{59}^ \circ }} \right) = \sec {59^ \circ }\)

Now you are provided with all the necessary information regarding trigonometric ratios of complementary angles. Practice more questions and master this concept.

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