Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

35. Why (1-t...

Question

35. Why { (1-tan x2) (1+tan x2) } = tan (4 - x2) ?

Open in App

Solution

Suggest Corrections

Similar questions

{tan}^{- 1} [\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}] =

Prove that $t a n^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) = \frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2} .$

\forall n \in N

, x

\in R

, the value of

{tan}^{- 1} [\frac{x}{1.2 + x^{2}}] + {tan}^{- 1} [\frac{x}{2.3 + x^{2}}] + \dots + {tan}^{- 1} [\frac{x}{n (n + 1) + x^{2}}]

, is

Q. Simplify

{tan}^{- 1} [\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}] =

Q. Solve the following equations for x:

(i)

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

(ii)

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

(iii)

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

(iv) 2

\tan^{- 1}

(

\sin

x

)

=

\tan

^{$-$ 1} (2

\sin

x

x \neq \frac{π}{2}

.

(v)

\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}

(vi)

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

Join BYJU'S Learning Program

Select...