Prove that: sin 270 ∘ + A = - cos A and tan 270 ∘ + A = - cot A .

( i ) . LHS = sin 270 ° + A = sin 360 ° - 90 ° - A = - sin 90 ∘ - A ∵ sin 360 ° - θ = - sin θ = - cos A ∵ sin 90 ∘ - θ = cos θ = RHS ( ii ) . LHS = tan 270 ° + A = tan 360 ° - 90 ° - A = - tan 90 ° - A ∵ tan 360 ° - θ = - tan θ = - cot A ∵ tan 90 ∘ - θ = cot θ = RHS

Prove that cos A + sin 270 ∘ + A - sin 270 ∘ - A + cos 180 ∘ + A = 0 .

Solving LHS cos A + sin 270 ∘ + A - sin 270 ∘ - A + cos 180 ∘ + A = cos A + - cos A - - cos A + - cos A ∵ sin 270 ∘ + A = - cos A , sin 270 ∘ - A = - cos A , cos 180 ∘ + A = - cos A = cos A - cos A + cos A - cos A = 0 = RHS

Prove that: sec 270 ∘ - A sec 90 ∘ - A - tan 270 ∘ - A tan 90 ∘ + A + 1 = 0 .

LHS = sec 270 ∘ - A sec 90 ∘ - A - tan 270 ∘ - A tan 90 ∘ + A + 1 = - cosec A · cosec A - cot A - cot A + 1 ∵ sec 270 ∘ - A = - cosec A , sec 90 ∘ - A = cosec A , tan 270 ∘ - A = cot A , tan 90 ∘ + A = - cot A = - cosec 2 A + cot 2 A + 1 = - cosec 2 A - cot 2 A + 1 = - 1 + 1 ∵ cosec 2 A - cot 2 A = 1 = 0 = RHS

Prove that: cot A + tan 180 ∘ + A + tan 90 ∘ + A + tan 360 ∘ - A = 0 .

Solving LHS = cot A + tan 180 ∘ + A + tan 90 ∘ + A + tan 360 ∘ - A = cot A + tan A + - cot A + - tan A ∵ tan 180 ∘ + θ = tan θ , tan 90 ∘ + θ = - cot θ , tan 360 ∘ - θ = - tan θ = 0 = RHS

E M B I B E

Plane Trigonometry Part 1>Chapter 5 - Trigonometrical Ratios of Angles of any Size and Sign>EXAMPLES X

EXAMPLES X

Author:S L Loney

S L Loney Mathematics Solutions for Exercise - EXAMPLES X

Simple step-by-step solutions to EXAMPLES X questions of Trigonometrical Ratios of Angles of any Size and Sign from Plane Trigonometry Part 1. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.