If tan x-t then tan 2 x- 2 x-a 1- t 1- tb 1- t 1- tc 2 t 1- td 2 t 1- t

(a) 1+t1 t tan #160;2x #160;+ #160;sec #160;2x=2 #160;tan #160;x1 tan2 #160;x+1+tan2 #160;x1 tan2 #160;x #160; #160; #160; #160; #1 ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If tan x=t th...

Question

If $\tan x = t$ then $\tan 2 x + \sec 2 x =$

(a) $\frac{1 + t}{1 - t}$

(b) $\frac{1 - t}{1 + t}$

(c) $\frac{2 t}{1 - t}$

(d) $\frac{2 t}{1 + t}$

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Solution

(a) $\frac{1 + t}{1 - t}$

$\tan 2 x + \sec 2 x = \frac{2 \tan x}{1 - \tan^{2} x} + \frac{1 + \tan^{2} x}{1 - \tan^{2} x} = \frac{2 \tan x + 1 + \tan^{2} x}{1 - \tan^{2} x} = \frac{{(1 + \tan x)}^{2}}{1 - \tan^{2} x} = \frac{(1 + \tan x) (1 + \tan x)}{(1 + \tan x) (1 - \tan x)} = \frac{1 + \tan x}{1 - \tan x} = \frac{1 + t}{1 - t} [\tan x = t (given)]$

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If tan x=t then tan 2x + sec 2x= (a) 1+t1-t (b) 1-t1+t (c) 2t1-t (d) 2t1+t

(a) 1+t1-t tan 2x + sec 2x=2 tan x1-tan2 x+1+tan2 x1-tan2 x =2tan x+1+tan2 x1-tan2x =1+tan x21-tan2 x =1+tan x1+tan x1+tan x1-tan x =1+tan x1-tan x =1+t1-t tan x=t given

If $\tan x = t$ then $\tan 2 x + \sec 2 x =$

(a) $\frac{1 + t}{1 - t}$

(b) $\frac{1 - t}{1 + t}$

(c) $\frac{2 t}{1 - t}$

(d) $\frac{2 t}{1 + t}$