Find the value of 1-tan2 x - 1-sin2 x

The correct option is C 1 Given, 1tan^2 x 1sin^2 x =^2x cosec^2x =^2x (1+^2x) .............[Since #160;cosec^2x=1+^2x ] = 1 ...

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

Trigonometric Identity- 2

Find the valu...

Question

Find the value of $\frac{1}{t a n^{2} x} - \frac{1}{s i n^{2} x}$

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

L t x \to 0 \frac{\sqrt[3]{1 + {sin}^{2} x} - \sqrt[4]{(1 - 2 tan x)}}{sin x + {tan}^{2} x}

\frac{1 + tan x + {tan}^{2} x + . . . . + t a n^{n} x + . . .}{1 - tan x + t a n^{2} x - . . . + {(- 1)}^{n} {tan}^{n} x + . . .} = 1 + sin 2 x

\frac{1 + sin x + cos x + sin 2 x + cos 2 x}{tan 2 x} = 0

Prove that

(i) $\frac{sin 2 x}{1 - cos 2 x} = cot x$

(ii) $\frac{1 - cos2x}{1 + cos 2 x} = {tan}^{2} x$

2 sin 2 x - cos 2 x = 1, x \neq (2 n + 1) \frac{π}{2}

sin 2 x + cos 2 x

\frac{1}{5}

sin 2 x + cos 2 x = \frac{1 + 2 tan x - {tan}^{2} x}{1 + {tan}^{2} x}

Join BYJU'S Learning Program