Prove that 1 tan x + cot x = sin x · cos x .

Given that, the trigonometric function is LHS = 1 tan x + cot x We know that, cot x = 1 tan x ⇒ LHS = 1 tan x + 1 tan x = tan x 1 + tan 2 x We know that, sec 2 θ = 1 + tan 2 θ ⇒ LHS = tan x sec 2 x We know that, tan θ = sin θ cos θ and secθ = 1 cosθ . ⇒ LHS = sin x cos x 1 cos 2 x = sin x cos x cos 2 x = sin x · cos x = RHS Hence, proved.

Prove that sec 2 x + sec x · tan x ≡ 1 1 - sin x .

Given that, the trigonometric function is LHS = sec 2 x + sec x tan x = sec x sec x + tan x We know that, sec θ = 1 cos θ and tan θ = sin θ cos θ ⇒ LHS = 1 cos x 1 cos x + sin x cos x = 1 + sin x cos 2 x We know that, sin 2 θ + cos 2 θ = 1 ⇒ LHS = 1 + sin x 1 - sin 2 x = 1 + sin x 1 - sin x 1 + sin x = 1 1 - sin x = RHS Hence, proved.

Prove that 1 - cos 2 x sec 2 x - 1 = 1 - sin 2 x .

Given that, the trigonometric function is LHS = 1 - cos 2 x sec 2 x - 1 We know that, sin 2 θ + cos 2 θ = 1 ⇒ LHS = sin 2 x sec 2 x - 1 We know that, sec 2 θ = 1 + tan 2 θ ⇒ LHS = sin 2 x tan 2 x = sin 2 x cos 2 x sin 2 x = cos 2 x = 1 - sin 2 x = RHS Hence, proved.

Prove that 1 + tan 2 x tan x = sec x · cosec x .

Given that, the trigonometric function is LHS = 1 + tan 2 x tan x . We know that, sec 2 θ = 1 + tan 2 θ ⇒ LHS = sec 2 x tan x = 1 cos 2 x cos x sin x = 1 cos x · sin x = sec x · cosecx = RHS Hence, proved.

Prove that sin x 1 - cos 2 x = cosec x .

Given that, the trigonometric function is LHS = sin x 1 - cos 2 x . We know that, sin 2 θ + cos 2 θ = 1 ⇒ LHS = sin x sin 2 x = 1 sin x = cosec x = RHS Hence, proved.

Prove that 1 + sin x 1 - sin x = ( tan x + sec x ) 2 .

Given that, the trigonometric function is RHS = ( tan x + sec x ) 2 = sin x cos x + 1 cos x 2 = 1 + sin x 2 cos 2 x . We know that, sin 2 θ + cos 2 θ = 1 ⇒ RHS = 1 + sin x 2 1 - sin 2 x = 1 + sin x 2 1 - sin x 1 + sin x = 1 + sin x 1 - sin x = LHS Hence, proved.

Prove that 1 1 + cos x + 1 1 - cos x ≡ 2 cosec 2 x .

Given that, the trigonometric function is LHS = 1 1 + cos x + 1 1 - cos x = 1 - cos x + 1 + cos x 1 + cos x 1 - cos x = 2 1 - cos 2 x . We know that, sin 2 θ + cos 2 θ = 1 ⇒ LHS = 2 sin 2 x = 2 cosec 2 x = RHS Hence, proved.

Prove that cos x 1 + sin x + cos x 1 - sin x ≡ 2 sec x .

Given that, the trigonometric function is LHS = cos x 1 + sin x + cos x 1 - sin x = cos x 1 - sin x + 1 + sin x 1 + sin x 1 - sin x = 2 cos x 1 - sin 2 x . We know that, sin 2 θ + cos 2 θ = 1 ⇒ LHS = 2 cos x cos 2 x = 2 cos x = 2 sec x = RHS Hence, proved.

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>EXERCISE 3A

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 2: EXERCISE 3A

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 2: EXERCISE 3A

Attempt the free practice questions on Chapter 3: Trigonometry, Exercise 2: EXERCISE 3A with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.