If ( 4 +x )+ ( 4 -x )= 2x, then:

MCQ

If $\tan\Big(\frac{\pi}{4}+\text{x}\Big)+\tan\Big(\frac{\pi}{4}-\text{x}\Big)=\lambda\sec2\text{x},$ then:

A
$3$
B
$4$
C
$1$
✓
$2$

✓

Answer

Correct option: D.

$2$

Given:
$\tan\big(\frac{\pi}{4}+\text{x}+\tan\big(\frac{\pi}{4}-\text{x}\big)=\lambda\sec2\text{x}$
$\Rightarrow\frac{\tan\frac{\pi}{4}+\tan\text{x}}{1-\tan\frac{\pi}{4}\times\tan\text{x}}+\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\times\tan\text{x}}=\lambda2\text{x}$
$\Rightarrow\frac{1+\tan\text{x}}{1-\tan\text{x}}+\frac{1-\tan\text{x}}{1+\tan\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow\frac{(1+\tan\text{x})^2+(1-\tan\text{x})^2}{(1-\tan\text{x})(1+\tan\text{x})}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2(1+\tan^2\text{x})}{1-\tan^2\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2\sec^2\text{x}}{1-\tan^2\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2}{\cos^2(1-\tan^2\text{x})}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2}{\cos^2\text{x}\Big(1-\frac{\sin^2\text{x}}{\cos^2\text{x}}\Big)}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2}{\cos^2\text{x}-\sin^2\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2}{\cos2\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow2\sec2\text{x}=\lambda\sec2\text{x}$
$\Rightarrow2=\lambda$
$\therefore\lambda=2$

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