Question

tan(\pi \div 4 + x \div 2)
​

Answer 1

Answer:

$sec \ x \ + \ tan \ x$

Step-by-step explanation:

Given :

To find the value of :

$tan(\frac{\pi}{4}+\frac{x}{2})$

Solution :

We know that,

π = 180°,

i.e,  $\frac{\pi}{4} = \frac{180}{4}$ = 45°

We know that,

tan 45° = 1,

Identities :

$tan \ A = \frac{sin \ A}{cos \ A}$

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

$cos \ 2A =2cos^2 \ A - 1 = 1 \ - \ 2sin^2 \ A=cos^2 \ A - sin^2 \ A$

$tan^2 \ A + 1 = sec^2 \ A$

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$tan(\frac{\pi}{4}+\frac{x}{2})$

⇒ $tan (\frac{\pi}{4}+\frac{x}{2}) =\frac{tan \frac{\pi}{4} \ + \ \tan \ \frac{x}{2}}{1 \ - \ tan \ \frac{\pi}{4} \ tan \ \frac{x}{2}}$

⇒ $tan (\frac{\pi}{4}+\frac{x}{2}) =\frac{1+ \ tan \ \frac{x}{2}}{1 \ - \ tan \ \frac{x}{2}}$

multiplying & dividing by $\frac{1 \ + \ tan \ \frac{x}{2}}{1 \ + \ tan \ \frac{x}{2}}$,

We get,

⇒ $tan (\frac{\pi}{4}+\frac{x}{2}) =\frac{1 \ + \ tan \ \frac{x}{2}}{1 \ - \ tan \ \frac{x}{2}} \times \frac{1 \ + \ tan \ \frac{x}{2}}{1 \ + \ tan \ \frac{x}{2}}$

⇒ $tan (\frac{\pi}{4}+\frac{x}{2}) =\frac{(1 \ + \ tan \ \frac{x}{2})^2}{1 \ - \ tan^2 \ \frac{x}{2}}$

⇒ $tan (\frac{\pi}{4}+\frac{x}{2}) =\frac{1 \ + \ tan^2 \ \frac{x}{2} \ + \ 2 \ tan \ \frac{x}{2}}{\frac{cos^2 \ \frac{x}{2} - sin^2 \ \frac{x}{2}}{cos^2 \ \frac{x}{2}}}$

⇒ $tan (\frac{\pi}{4}+\frac{x}{2}) =\frac{sec^2 \ \frac{x}{2} \ + \ 2 \ (\frac{sin \ \frac{x}{2}}{cos \frac{x}{2}})}{\frac{cos^2 \ \frac{x}{2} - sin^2 \ \frac{x}{2}}{cos^2 \ \frac{x}{2}}}$

⇒ $tan (\frac{\pi}{4}+\frac{x}{2}) =\frac{\frac{1}{cos^2 \ \frac{x}{2}} \ + \ (\frac{2sin \ \frac{x}{2}}{cos \frac{x}{2}})}{\frac{cos^2 \ \frac{x}{2} - sin^2 \ \frac{x}{2}}{cos^2 \ \frac{x}{2}}}$

⇒ $tan (\frac{\pi}{4}+\frac{x}{2}) =\frac{\frac{1 \ + \ 2 \ sin \ \frac{x}{2} \ cos \ \frac{x}{2}}{cos^2 \frac{x}{2}}}{\frac{cos \ x}{cos^2 \ \frac{x}{2}}}$

⇒ $tan (\frac{\pi}{4}+\frac{x}{2})=\frac{\frac{1 \ + \ sin \ x}{cos^2 \frac{x}{2}}}{\frac{cos \ x}{cos^2 \ \frac{x}{2}}}$

⇒ $\frac{1 \ + \ sin \ x}{cos \ x}$

⇒ $\frac{1}{cos \ x} + \frac{sin \ x}{cos \ x} = sec \ x \ + \ tan \ x$

Answer 2

Answer:

drpok gli ka kutta........