Question

Sin (90-theta )cos (90-theta) = tan theta / 1 + cot^2 (90 - theta)

Answer 1

Step-by-step explanation:

LHS

sin(90-θ)cos(90-θ)

=cosθsinθ

RHS

tanθ/1+cot²(90-θ)

=tanθ/1+tan²θ

=tanθ/sec²θ

=sinθ/cosθ × cos²θ

=sinθcosθ

LHS=RHS

Answer 2

$\huge{\underline{\underline{\purple{♡Answer→}}}}$

$\sin(90 - \theta) \cos(90 - \theta) = \frac{ \tan(\theta) }{1 + { \cot^{2} (90 -\theta ) }}$

LHS →

$= \sin(90 - \theta) \cos(90 - \theta)$

$= \cos(\theta) \sin(\theta)$

RHS →

$= \frac{ \tan(\theta) }{1 + { \cot^{2} (90 -\theta ) }}$

$= \frac{ \tan(\theta) }{1 + { \tan^{2}(\theta) } }$

$= \frac{tan \: \theta}{ {sec}^{2} \theta}$

$= \frac{sin \: \theta}{cos \: \theta} \times {cos}^{2} \theta$

$= \sin(\theta) \cos(\theta) \:$

__________________________

LHS = RHS

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