Prove sin theta -8203- -sin 90 - theta- - cos nbsp-theta - cos - 90 - theta -8203- sec nbsp-theta -8203- cosec theta -8203-

Dear student, Taking #160;L.H.Ssin #952;sin90 #952;+cos #952;cos90 #952;=sin #952;cos #952;+cos #952;sin #952;=sin2 #952;+cos2 #952;cos ...

You visited us 9 times! Enjoying our articles? Unlock Full Access!

Standard Values of Trigonometric Ratios

Prove sin the...

Question

Prove sin theta / (sin 90 - theta) + cos theta / cos ( 90 - theta )
= sec theta * cosec theta

Open in App

Solution

Suggest Corrections

Similar questions

\frac{sin θ cos (90^{\circ} - θ) cos θ}{cosec (90^{\circ} - θ)}

-

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

Find the value of
$s i n θ$ $c o s θ$ [ $s i n (90^{\circ} - θ)$ $c o s e c θ$ + $c o s (90^{\circ} - θ)$ $s e c θ$ ]

$s i n θ$ $c o s θ$ [ $s i n (90^{\circ} - θ)$ $c o s e c θ$ + $c o s (90^{\circ} - θ)$ $s e c θ$ ]

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)}

\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{cos (90 ° + θ) cosec (270 ° + θ) cos (180 ° + θ)}{sec (- θ) sin (270 ° + θ) sin (360 ° - θ)} =

Join BYJU'S Learning Program