Question

Prove that √ (1+Sin A/1-SinA) = SecA + Tan A​

Answer 1

Answer:

√ (1+Sin A/1-SinA) = SecA + Tan A​

Multiplying 1+ sinA in LHS under roots

= √ [(1+SinA)²/(1-SinA)(1 + sinA)]

= √ [(1+SinA)²/(cos²A)]                               { Sin²A + cos²A =1}

= 1 + sinA / cosA

= 1/cosA + sinA/cosA

= secA + tanA

∴ LHS = RHS

HOPE THIS HELPS YOU...........

PLZ MARK AS BRAINLEST..........

Answer 2

To Prove:

$\displaystyle\bf\sqrt{\Bigg(\dfrac{1+sin A}{1-sin A}\Bigg)}=sec A+tan A$

Solution:-

$\displaystyle\Longrightarrow \sf\sqrt{\Bigg(\dfrac{1+\sin A}{1-\sin A}\Bigg)}=\sec A+\tan A$

Now rationalise it!

$\displaystyle\sf\Longrightarrow\sqrt{\Bigg[\dfrac{1+\sin A}{1-\sin A}\times\dfrac{1+\sin A}{1+\sin A}\Bigg]}=\sec A+\tan A$

Now multiply in LHS

$\displaystyle\sf\Longrightarrow\sqrt{\Bigg[\dfrac{\big(1+\sin A\big)\big(1+\sin A\big)}{\big(1-\sin A\big)\big(1+\sin A\big)}\Bigg]}=\sec A+\tan A$

$\displaystyle\sf\Longrightarrow\sqrt{\Bigg[\dfrac{\big(1+\sin A\big)^{2}}{\big(1-\sin A\big)\big(1+\sin A\big)}\Bigg]}=\sec A+\tan A$

Now apply Identity:

(A+B)(A-B)=A²-B²

$\displaystyle\sf\Longrightarrow\sqrt{\Bigg[\dfrac{\big(1+\sin A\big)^{2}}{\big(1\big)^2-\big(\sin A\big)^2}\Bigg]}=\sec A+\tan A$

$\displaystyle\sf\Longrightarrow\sqrt{\Bigg[\dfrac{\big(1+\sin A\big)^{2}}{\big(1-\sin^2 A\big)}\Bigg]}=\sec A+\tan A$

Now applying identity:

• 1-sin²A=Cos²A

$\displaystyle\sf\Longrightarrow\sqrt{\Bigg[\dfrac{\big(1+\sin A\big)^{2}}{\big(\cos^2A)}\Bigg]}=\sec A+\tan A$

$\displaystyle\sf\Longrightarrow\sqrt{\Bigg(\dfrac{1+\sin A}{\cos A}\Bigg)^2}=\sec A+\tan A$

$\displaystyle\sf\Longrightarrow{\Bigg(\dfrac{1+\sin A}{\cos A}\Bigg)^{2\times\frac{1}{2}}}=\sec A+\tan A$

$\displaystyle\sf\Longrightarrow{\Bigg(\dfrac{1+\sin A}{\cos A}\Bigg)^{\not 2\times\frac{1}{\not2}}}=\sec A+\tan A$

$\displaystyle\sf\Longrightarrow{\Bigg(\dfrac{1+\sin A}{\cos A}\Bigg)^1=\sec A+\tan A$

Now taking denominator for both numerator

$\displaystyle\sf\Longrightarrow{\Bigg(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\Bigg)=\sec A+\tan A$

Applying formula:

$\bullet\quad\sf\displaystyle\frac{1}{\cos A}=\sec A\\\\\\\bullet\quad\displaystyle\dfrac{\sin A}{\cos A}=\tan A$

$\displaystyle\sf\Longrightarrow\sec A+\tan A=\sec A+\tan A$

$\large\underline{\boxed{\sf{\red{ Hence\; Proved}}}}\bigstar$