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
= cos2 θ + sin2 θ
= 1 = R.H.S.
(ii) L.H.S = tan θ/ tan (90o – θ) + sin (90o – θ)/ cos θ
= tan θ/ cot θ + cos θ/ cos θ
= tan θ/ (1/tan θ) + 1
= tan2 θ + 1 = sec2 θ = R.H.S.
(iii) L.H.S. = (cos (90o – θ) cos θ)/ tan θ + cos2 (90o – θ)
= (sin θ cos θ)/ tan θ + sin2 θ
= (sin θ cos θ)/ (sin θ/ cos θ) + sin2 θ
= cos2 θ + sin2 θ
= 1 = R.H.S.