If \( \tan\left(\frac{\pi}{4}+x\right)+\tan\left(\frac{\pi}{4}-x\right)=\lambda\sec2x \), Find \( \lambda \)
Question
If
\[ \tan\left(\frac{\pi}{4}+x\right) + \tan\left(\frac{\pi}{4}-x\right) = \lambda\sec2x, \]
then \(\lambda\) is
(a) 3
(b) 4
(c) 1
(d) 2
Solution
Let
\[ A=\frac{\pi}{4}+x,\qquad B=\frac{\pi}{4}-x \]
Using the identity
\[ \tan A+\tan B = \frac{\sin(A+B)} {\cos A\cos B} \]
Since
\[ A+B=\frac{\pi}{2} \]
we get
\[ \tan A+\tan B = \frac{\sin\frac{\pi}{2}} {\cos A\cos B} = \frac{1} {\cos A\cos B} \]
Now,
\[ 2\cos A\cos B = \cos(A+B)+\cos(A-B) \]
\[ = \cos\frac{\pi}{2} + \cos2x = \cos2x \]
Therefore,
\[ \cos A\cos B = \frac{\cos2x}{2} \]
Hence,
\[ \tan\left(\frac{\pi}{4}+x\right) + \tan\left(\frac{\pi}{4}-x\right) = \frac{1}{\cos2x/2} = \frac{2}{\cos2x} \]
\[ = 2\sec2x \]
Comparing with
\[ \lambda\sec2x, \]
we obtain
\[ \lambda=2 \]
Final Answer
\[ \boxed{\lambda=2} \]
Hence, the correct option is (d) 2.