Prove that- - 1 - - - - - - 2- 1 - - - - - - 2- - 2-

[1+θ (θ+π2)][1+θ+(θ+π2)] =[1+θ+θ][1+θ θ] =(1+θ)^2 ^2θ =1+2θ+^2θ ^2θ =1+2θ 1 =2θ ...

250

You visited us 250 times! Enjoying our articles? Unlock Full Access!

Trigonometric Identities

Prove that: ...

Question

Prove that: $[1 + cot θ - sec (θ + \frac{π}{2})] [1 + cot θ + sec (θ + \frac{π}{2})] = 2 cot θ$

Open in App

Solution

Suggest Corrections

Similar questions

\frac{cot θ}{csc θ + 1} + \frac{csc θ + 1}{cot θ} = 2 sec θ

$i f \frac{1}{{cot}^{2} θ} + {cot}^{2} θ = 2$ then prove that $cot θ + \frac{1}{cot θ} = 2$

\sqrt{\frac{sec θ - 1}{sec θ + 1}} + \sqrt{\frac{sec θ + 1}{sec θ - 1}} = 2 csc θ

\frac{\cos (2 π + θ) cosec (2 π + θ) \tan (π / 2 + θ)}{\sec (π / 2 + θ) cosθ \cot (π + θ)} = 1

\frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = 2

\frac{\sin (180 ° + θ) \cos (90 ° + θ) \tan (270 ° - θ) \cot (360 ° - θ)}{\sin (360 ° - θ) \cos (360 ° + θ) cosec (- θ) \sin (270 ° + θ)} = 1

\{1 + cotθ - \sec (\frac{π}{2} + θ)\} \{1 + cotθ + \sec (\frac{π}{2} + θ)\} = 2 cotθ

\frac{\tan (90 ° - θ) \sec (180 ° - θ) \sin (- θ)}{\sin (180 ° + θ) \cot (360 ° - θ) cosec (90 ° - θ)} = 1

Prove that : $\frac{t a n θ}{s e c θ - 1}$ - $\frac{t a n θ}{s e c θ + 1}$ = 2 cot $θ$

Join BYJU'S Learning Program