Prove- - 2-cosec2-tan-

LHS = √(sec^2θ + cosec^2θ) = √((1 + tan^2θ) + (1 + cot^2θ)) = √(tan^2θ + cot^2θ + 2 tanθ cotθ) = √((tanθ + cotθ)^2) = ta ...

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

Trigonometric Identities

Prove: √ 2θ+c...

Question

Prove: $\sqrt{s e c^{2} θ + c o s e c^{2} θ}$ = $t a n θ + c o t θ$

Open in App

Solution

Suggest Corrections

Similar questions

\sqrt{s e c^{2} θ + c o s e c^{2} θ} = t a n θ + c o t θ

\sqrt{{sec}^{2} θ + c o s e c^{2} θ} = tan θ + cot θ

\sqrt{s e c^{2} θ + \cos e c^{2} θ} = \tan θ + c o t θ

If $θ$ is acute angle, and $t a n θ$ = $\frac{1}{\sqrt{7}}$ , then $\frac{c o s e c^{2} θ - s e c^{2} θ}{c o s e c^{2} θ + s e c^{2} θ} =$

\frac{1}{1 - \sin θ} + \frac{1}{1 + \sin θ} = 2 \sec^{2} θ

\frac{\tan θ}{s e c θ + 1} = \frac{s e c θ - 1}{\tan θ}

\frac{\tan^{3} θ - 1}{\tan θ - 1} = \sec^{2} θ + \tan θ

\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{\sin θ - \tan θ}

Join BYJU'S Learning Program