Prove that - cos - sin90- - - - sin - cos90- - - - 1-

LHS = cos θ #160; sin ( 90^∘ #160; θ ) + sin θ #160; cos ( 90^∘ #160; θ ) ⇒ LHS = cos θ cos θ sin θ sin θ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Domain

Prove that : ...

Question

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

Open in App

Solution

Suggest Corrections

Similar questions

\frac{s i n θ c o s (90^{\circ} - θ) c o s θ}{s i n (90^{\circ} - θ)} + \frac{c o s θ s i n (90^{\circ} - θ) s i n θ}{c o s (90^{\circ} - θ)} = 1

sin θ cos (90 - θ) + cos θ sin (90 - θ) =

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{3}{2}

Join BYJU'S Learning Program