Prove the following:

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Prove the following:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
(ii) \frac { tan\theta }{ tan({ 90 }^{ O }-\theta ) } +\frac { sin({ 90 }^{ O }-\theta ) }{ cos\theta } ={ sec }^{ 2 }\theta
(iii) \frac { cos({ 90 }^{ O }-\theta )cos\theta }{ tan\theta } +{ cos }^{ 2 }({ 90 }^{ O }-\theta )=1
(iv) sin (90° – θ) cos (90° – θ) = \frac { tan\theta }{ { 1+tan }^{ 2 }\theta }

Solution:

(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
L.H.S. = cos θ sin (90° – θ) + sin θ cos (90° – θ)
= cos θ . cos θ + sin θ . sin θ
= cos2 θ + sin2 θ = 1 = R.H.S.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q11.1

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q11.2

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 Q11.3

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