Name: Formulas for Trigonometric Functions
Uploaded: 2019-07-19
Duration: 21 min 44 s
Description: In this video, you will learn the formulas for trigonometric functions: Sum/Difference, Double/Half-Angle, Prod-to-Sum/Sum-to-Prod. So let's watch this video to know more.

Question

In a

△ A B C

if

C = 90 °

, then prove that

(\frac{a^{2} + b^{2}}{a^{2} - b^{2}}) \sin (A - B) = 1

.

Accepted Answer

Given that, $∆ A B C$ , $∠ C = 90 °$

Then, $A + B + C = 180 °$

$\Rightarrow A + B = 90 °$ ...(i)

From sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

To prove, $(\frac{a^{2} + b^{2}}{a^{2} - b^{2}}) \sin (A - B) = 1$

$LHS = (\frac{a^{2} + b^{2}}{a^{2} - b^{2}}) \sin (A - B) = 1$

$= \frac{(k \sin A)^{2} + (k \sin B)^{2}}{(k \sin A)^{2} - (k \sin B)^{2}} \sin (A - B)$

$= \frac{\sin^{2} A + \sin^{2} B}{\sin^{2} A - \sin^{2} B} \sin (A - B)$ [Cancelling $k$ from numerator and denominator]

$= \frac{\frac{1 - \cos 2 A}{2} + \frac{1 - \cos 2 B}{2}}{\frac{1 - \cos 2 A}{2} - \frac{1 - \cos 2 B}{2}} \sin (A - B)$

$= \frac{1 - \cos 2 A + 1 - \cos 2 B}{1 - \cos 2 A - 1 + \cos 2 B} \sin (A - B)$

$= \frac{2 - \cos 2 A - \cos 2 B}{\cos 2 B - \cos 2 A} \sin (A - B)$

$= \frac{2 - (\cos 2 A + \cos 2 B)}{\cos 2 B - \cos 2 A} \sin (A - B)$

$= \frac{2 - (2 \cos \frac{(2 A + 2 B)}{2} \cos \frac{(2 A - 2 B)}{2})}{2 \sin \frac{(2 B + 2 A)}{2} \sin \frac{(2 A - 2 B)}{2}} \sin (A - B)$

$= \frac{2 - (2 \cos (A + B) \cos (A - B))}{2 \sin (A + B) \sin (A - B)} \sin (A - B)$

$= \frac{2 - (2 \cos (A + B) \cos (A - B))}{2 \sin (A + B) \sin (A - B)} \sin (A - B)$

$= \frac{2 - (2 \cos (A + B) \cos (A - B))}{2 \sin (A + B)}$

From (i), we get

$= \frac{2 - (2 \cos 90 ° \cos (A - B))}{2 \sin 90 °}$

$= \frac{2 - 0}{2}$

$= 1$

$= RHS$

Hence, proved.

Exercise 10 (a)

Telangana Board Mathematics Solutions for Exercise 10 (a)

Questions from Exercise 10 (a) with Hints & Solutions

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