Question

প্রমাণ কর: $\sqrt{\frac{1-sinA}{1+sinA}}$$=secA-tanA$

Answer 1

Left Hand Side
$=\sqrt{\frac{1-sinA}{1+sinA}}$

$=\sqrt{\frac{(1-sinA)(1-sinA)}{(1+sinA)(1-sinA)}}$
[ লব ও হর’কে $\sqrt{(1-sinA)}$ দ্বারা গুণ করে ]

$=\sqrt{\frac{(1-sinA)^2}{1^2-sin^2A}}$

$=\sqrt{\frac{(1-sinA)^2}{1-sin^2A}}$

$=\sqrt{\frac{(1-sinA)^2}{cos^2A}}$

$=\frac{1-sinA}{cosA}$

$=\frac{1}{cosA}-\frac{sinA}{cosA}$

$=secA-tanA$

$=$ Right Hand Side

[Proved]

Comments

Rules Applied:

• $a^2-b^2=(a+b)(a-b)$
• $1-sin^2\theta=cos^2\theta$
• $sec\theta=\frac{1}{cos\theta}$
• $tan\theta=\frac{sin\theta}{cos\theta}$