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Question Number 106118 by DeepakMahato last updated on 02/Aug/20

Commented by DeepakMahato last updated on 02/Aug/20

$P l e a s e S o l v e . . .$
	Answered by nimnim last updated on 02/Aug/20

	$A + B = 90 \Rightarrow A = 90 - B o r B = 90 - A$ $L H S : \sqrt{\frac{t a n A t a n B + t a n A c o t B}{s i n A s e c B} - \frac{{s i n}^{2} B}{{c o s}^{2} A}}$ $= \sqrt{\frac{t a n A t a n (90 - A) + t a n A c o t (90 - A)}{s i n A s e c (90 - A)} - \frac{{s i n}^{2} (90 - A)}{{c o s}^{2} A}}$ $= \sqrt{\frac{t a n A . c o t A + t a n A t a n A}{s i n A . c o s e c A} - \frac{{c o s}^{2} A}{{c o s}^{2} A}}$ $= \sqrt{\frac{1 + {t a n}^{2} A}{1} - 1} = \sqrt{{s e c}^{2} A - 1}$ $= \sqrt{{t a n}^{2} A} = t a n A (R H S)$
	Commented by DeepakMahato last updated on 02/Aug/20
	Thank You Once Again...
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