Question

Tan(π÷4 + A÷2) = 1 ÷ secA - tanA

Answer 1

To Prove: $\tan(\frac{\Pi }{4}+\frac{A}{2})$ = $\frac{1}{\sec A-\tan A}$

Consider LHS :

$\tan(\frac{\Pi }{4}+\frac{A}{2})$

By using the identity $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$=\frac{\tan \frac{\Pi }{4}+\tan \frac{A}{2}}{1-\tan \frac{\Pi }{4} \tan \frac{A}{2}}$

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

Converting in Sine and Cosine form, we get

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

=$\frac{\cos\frac{A}{2}+ \sin \frac{A}{2}}{\cos\frac{A}{2}- \sin \frac{A}{2}} \times \frac{\cos\frac{A}{2}+ \sin \frac{A}{2}}{\cos\frac{A}{2}+ \sin \frac{A}{2}}$

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

=$\frac{(\cos^{2}\frac{A}{2}+ \sin^{2} \frac{A}{2}+2 \sin \frac{A}{2} \cos\frac{A}{2})}{\cos{A}}$

=$\frac{(1+\sin A)}{\cos{A}}$

=$\sec A + \tan A$

Now, consider RHS of the equation,

$\frac{1}{\sec A-\tan A}$

=$\frac{1}{\sec A-\tan A} \times \frac{\sec A + \tan A}{\sec A + \tan A}$

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

= $\sec A + \tan A$

Therefore, LHS = RHS.

Hence, proved.

Answer 2

HELLO DEAR,

TO PROVE :-

Tan(π/4 + A/2) = 1 /( secA - tanA)

WE KNOW:-

sin²A + cos²A = 1,

cos²A - sin²A = cos2A

2sinAcosA = sin2A

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

so, $\bold{tan(\Pi/4 + A/2) = \frac{tan\Pi/4 + tanA/2}{1 - tan\Pi/4*tanA/2}}$

[tanπ/4 = 1]

$\bold{\Rightarrow tan(\Pi/4 + A/2) = \frac{1 + tanA/2}{1 - tanA/2}}$

$\bold{\Rightarrow \frac{1 + \frac{sinA/2}{cosA/2}}{1 - \frac{sinA/2}{cosA/2}}}$

$\bold{\Rightarrow \frac{sinA/2 + cosA/2}{cosA/2 - sinA/2}}$

$\bold{\Rightarrow \frac{sinA/2 + cosA/2}{sinA/2 - cosA/2} * \frac{cosA/2 + sinA/2}{cosA/2 + sinA/2}}$

$\bold{\Rightarrow \frac{(sinA/2 + cosA/2)^2}{(cos^2A/2 - sin^2A/2)}}$

$\bold{\Rightarrow \frac{(sin^2A/2 + cos^2A/2) + 2sinA/2*cosA/2}{cos2*(\frac{A}{2})}}$

$\bold{\Rightarrow \frac{1 + sinA}{cosA}}$

$\bold{\Rightarrow}$ secA + tanA

$\bold{\Rightarrow \frac{secA + tanA}{sec^2A - tan^2A}}$

$\bold{\to\to\to\to\to\to\to\to\to\to\to\boxed{sin^2\Theta - tan^2\Theta = 1}}$

$\bold{\Rightarrow \frac{1}{secA - tanA} * \frac{secA + tanA}{secA + tanA}}$

$\bold{\therefore, \Rightarrow \frac{1}{secA - tanA}}$

$\mathbb{\ulcorner \star HENCE,\, L.H.S = R.H.S \star \urcorner}$

I HOPE ITS YOU DEAR,

THANKS