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6.7.6 Trigonometric Functions Short Questions (Question 15 – 18)


Question 15:
Prove the identity 2 cos 2 A + 1 = s e c 2 A

Solution:
LHS = 2 cos 2 A + 1 = 2 ( 2 cos 2 A − 1 ) + 1 ← cos 2 A = 2 cos 2 A − 1 = 2 2 cos 2 A = 1 cos 2 A = s e c 2 A = RHS ∴ Proven
 


Question 16:
Prove the identity 2 tan A 2 − s e c 2 A = tan 2 A

Solution:
LHS = 2 tan A 2 − s e c 2 A = 2 tan A 2 − ( tan 2 A + 1 ) ← tan 2 A + 1 = s e c 2 A = 2 tan A 1 − tan 2 A = tan 2 A = RHS ∴ Proven



Question 17:
Prove the identity tan x + cot x = 2 cos e c 2 x

Solution:
LHS = tan x + cot x = sin x cos x + cos x sin x = sin 2 x + cos 2 x cos x sin x = 1 cos x sin x ← sin 2 x + cos 2 x = 1 = 1 1 2 sin 2 x ← sin 2 x = 2 sin x cos x 1 2 sin 2 x = sin x cos x = 2 sin 2 x = 2 ( 1 sin 2 x ) = 2 cos e c 2 x = RHS ∴ Proven
 


Question 18:
Prove the identity cos x − sin 2 x cos 2 x + sin x − 1 = 1 tan x

Solution:
LHS = cos x − sin 2 x cos 2 x + sin x − 1 = cos x − 2 sin x cos x ( 1 − 2 sin 2 x ) + sin x − 1 ← cos 2 x = 1 − 2 sin 2 x = cos x ( 1 − 2 sin x ) sin x − 2 sin 2 x = cos x ( 1 − 2 sin x ) sin x ( 1 − 2 sin x ) = cos x sin x = cot x = 1 tan x = RHS ∴ Proven

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