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# 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)

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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

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1. 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.

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