Question

$tanA+sinA=a$ এবং $tanA-sinA=b$ হলে, প্রমাণ কর যে, $a^2-b^2=4\sqrt{ab}$

Answer 1

দেওয়া আছে,
$tanA+sinA=a$ এবং $tanA-sinA=b$

Left Hand Side
$=a^2-b^2$

$=(tanA+sinA)^2-(tanA-sinA)^2$

$=4 \; tanA \cdot sinA$

$=4 \sqrt{tan^2A \cdot sin^2A}$

$=4 \sqrt{tan^2A \cdot (1-cos^2A)}$

$=4 \sqrt{tan^2A-tan^2A \cdot cos^2A}$

$=4 \sqrt{tan^2A-\frac{sin^2A}{cos^2A} \cdot cos^2A}$

$=4 \sqrt{tan^2A-sin^2A}$

$=4 \sqrt{(tanA+sinA)(tanA-sinA)}$

$=4 \sqrt{ab}$
[ মান বসিয়ে ]

$=$ Right Hand Side

[Proved]

Comments

Rules Applied :

• $(a+b)^2-(a-b)^2=4ab$
• $sin^2\theta=1-cos^2\theta$
• $tan\theta=\frac{sin\theta}{cos\theta}$