Question

tan(π/4+x/2)+tan(π/4-x/2)=2secx

Answer 1

Answer:

$\red { tan\left( \frac{\pi}{4}+\frac{x}{2}\right) + tan\left( \frac{\pi}{4}-\frac{x}{2}\right)}\green {=2secx}$

Step-by-step explanation:

$LHS = tan\left( \frac{\pi}{4}+\frac{x}{2}\right) + tan\left( \frac{\pi}{4}-\frac{x}{2}\right)$

$Let \: \frac{x}{2} = A\: ---(1)$

$= tan\left( \frac{\pi}{4}+A\right) + tan\left( \frac{\pi}{4}-A\right)$

$= \frac{tan\frac{pi}{4}+tanA}{1-tan\frac{\pi}{4}}+\frac{tan\frac{pi}{4}+tanA}{1-tan\frac{\pi}{4}}$

$\boxed {\pink { tan(A+B)= \frac{tanA + tanB}{1 -tanAtanB}}}$

$\boxed {\orange { tan(A-B)= \frac{tanA - tanB}{1 +tanAtanB}}}$

$= \frac{ 1+ tanA}{1-tanA} + \frac{1-tanA}{1+tanA}$

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

$= \frac{2( 1 + tan^{2}A)}{1 - tan^{2}A}$

$\boxed { \pink { \since , (a+b)^{2}+(a-b)^{2} = 2(a^{2}+b^{2})}}$

$\boxed { \orange { (a-b)(a+b) = a^{2} - b^{2}}}$

$= \frac{ 2sec^{2}A}{1-tan^{2}A}$

/* Divide numerator and denominator by sec²A ,we get

$= \frac{2}{\frac{1}{sec^{2}A}- \frac{tan^{2}A}{sec^{2}A}}$

$= \frac{2}{ cos^{2}A - sin^{2}A}$

$= \frac{2}{cos2A}$

$\boxed {\blue { cos^{2}A - sin^{2}A = cos2A}}$

$= \frac{2}{cos2\times \frac{x}{2}}\:[From \:(1)]$

$= \frac{2}{cosx}$

$= 2sec\:x$

$= R.H.S$

Therefore.,

$\red { tan\left( \frac{\pi}{4}+\frac{x}{2}\right) + tan\left( \frac{\pi}{4}-\frac{x}{2}\right)}\green {=2secx}$

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