If tanx+π4=a then sec2x=

tanx+π4=a 1+tan x1-tan x=a tan x=a-1a+1 sec2x=1+tan2x =1+a-1a+12 =(a+1)2+(a-1)2(a+1)2 =2(a2+1)(a+1)2

If tanx+π4=a then sec2x=

tanx+π4=a 1+tan x1-tan x=a tan x=a-1a+1 sec2x=1+tan2x =1+a-1a+12 =(a+1)2+(a-1)2(a+1)2 =2(a2+1)(a+1)2

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If $\tan (x + \frac{π}{4}) = a$ then $\sec^{2} x =$

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$1 + a^{2}$

$\frac{4 a}{{(a + 1)}^{2}}$

$\frac{2 (a^{2} + 1)}{{(a + 1)}^{2}}$

${(\frac{a - 1}{a + 1})}^{2}$

answer is C.

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Detailed Solution

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

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If $\tan (x + \frac{π}{4}) = a$ then $\sec^{2} x =$