Sin 90- cos 90- 90-sin 2-

The correct option is B 1sin(90 – θ) = cos θ cos(90 – θ) = sin θ cot(90 – θ) = tan θ cosθ sinθ/tanθ + sin^2θ = cos^2θ + sin^2θ = 1 ...

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Trigonometric Ratios of Standard Angles

sin 90-θ cos ...

Question

$\frac{s i n (90 - θ) c o s (90 - θ)}{c o t (90 - θ)}$ + $s i n^{2} θ$ =

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$\frac{sin (90^{\circ} - θ) cos (90^{\circ} - θ)}{cot (90^{\circ} - θ)} + {sin}^{2} θ =$ ___

$\frac{s i n (90 - θ) c o s (90 - θ)}{c o t (90 - θ)}$ + $s i n^{2} θ$ =

Prove that:
$\frac{s i n (90 - θ) c o s (90 - θ)}{c o t (90 - θ)}$ + $s i n^{2} θ$ = 1.

sinθ \cos (90 ° - θ) + \sin (90 ° - θ) cosθ = 1

\frac{sinθ}{\cos (90 ° - θ)} + \frac{cosθ}{\sin (90 ° - θ)} = 2

\frac{sinθ \cos (90 ° - θ) cosθ}{\sin (90 ° - θ)} + \frac{cosθ \sin (90 ° - θ) sinθ}{\cos (90 ° - θ)} = 1

\frac{\cos (90 ° - θ) \sec (90 ° - θ) tanθ}{cosec (90 ° - θ) \sin (90 ° - θ) \cot (90 ° - θ)} + \frac{\tan (90 ° - θ)}{cotθ} = 2

\frac{\cos (90 ° - θ)}{1 + \sin (90 ° - θ)} + \frac{1 + \sin (90 ° - θ)}{\cos (90 ° - θ)} = 2 cosecθ

\frac{\sec (90 ° - θ) cosecθ - \tan (90 ° - θ) cotθ + \cos^{2} 25 ° + \cos^{2} 65 °}{3 \tan 27 ° \tan 63 °} = \frac{2}{3} [CBSE 2010]

cotθ \tan (90 ° - θ) - \sec (90 ° - θ) cosecθ + \sqrt{3} \tan 12 ° \tan 60 ° \tan 78 ° = 2 [CBSE 2010]

\frac{sec (90^{\circ} - θ)}{sin (90^{\circ} - θ)} \times \frac{cos θ}{tan (90^{\circ} - θ)} - sec θ

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