This ques is very important and has been taken from the

Book- ML Aggarwal

Board- ICSE

Publication- Avichal

Chapter- Trigonometric Identities

Chapter number-18

This ques has many sub questions.

Prove that following: (i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1 (ii) tan θ/ tan (90o – θ) + sin (90o – θ)/ cos θ = sec2 θ (iii) (cos (90o – θ) cos θ)/ tan θ + cos2 (90o – θ) = 1 (iv) sin (90o – θ) cos (90o – θ) = tan θ/ (1 + tan2 θ)

Class10, ML Aggarwal, ques no 11, ICSE board

Solution:(i) L.H.S. = cos θ sin (90° – θ) + sin θ cos (90° – θ)

= cos θ x cos θ + sin θ x sin θ

= cos

^{2}θ + sin^{2}θ= 1 = R.H.S.

(ii) L.H.S = tan θ/ tan (90

^{o}– θ) + sin (90^{o}– θ)/ cos θ= tan θ/ cot θ + cos θ/ cos θ

= tan θ/ (1/tan θ) + 1

= tan

^{2}θ + 1 = sec^{2}θ = R.H.S.(iii) L.H.S. = (cos (90

^{o}– θ) cos θ)/ tan θ + cos^{2}(90^{o}– θ)= (sin θ cos θ)/ tan θ + sin

^{2}θ= (sin θ cos θ)/ (sin θ/ cos θ) + sin

^{2}θ= cos

^{2}θ + sin^{2}θ= 1 = R.H.S.