If tan^2 α tan^2 β+tan^2 β tan^2 γ+tan^2 γ tan^2 α +2tan^2 α tan^2 β tan^2 γ=1, then the value of sin^2 α+sin^2 β+sin^2 γ is


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Question Number 46483 by peter frank last updated on 27/Oct/18

$If \tan^{2} α \tan^{2} β + \tan^{2} β \tan^{2} γ + \tan^{2} γ \tan^{2} α$ $+ {2 \tan}^{2} α \tan^{2} β \tan^{2} γ = 1, then the value$ $of \sin^{2} α + \sin^{2} β + \sin^{2} γ is$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Oct/18

	$t a n α = a t a n β = b t a n γ = c$ $a^{2} b^{2} + b^{2} c^{2} + a^{2} c^{2} + 2 a^{2} b^{2} c^{2} = 1 \leftarrow (1)$ $t o f i n d$ $\frac{a^{2}}{a^{2} + 1} + \frac{b^{2}}{b^{2} + 1} + \frac{c^{2}}{c^{2} + 1} = ?$ $n o w f r o m (1) w e g e t$ $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} + 2 = \frac{1}{a^{2} b^{2} c^{2}}$ $n o w s a y$ $k = \frac{a^{2}}{a^{2} + 1} + \frac{b^{2}}{b^{2} + 1} + \frac{c^{2}}{c^{2} + 1}$ $k = \frac{a^{2} (b^{2} + 1) (c^{2} + 1) + b^{2} (a^{2} + 1) (c^{2} + 1) + c^{2} (a^{2} + 1) (b^{2} + 1)}{(a^{2} + 1) (b^{2} + 1) (c^{2} + 1)}$ $\frac{a^{2} (b^{2} c^{2} + b^{2} + c^{2} + 1) + b^{2} (a^{2} c^{2} + a^{2} + c^{2} + 1) + c^{2} (a^{2} b^{2} + a^{2} + b^{2} + 1)}{a^{2} (b^{2} c^{2} + b^{2} + c^{2} + 1) + b^{2} c^{2} + b^{2} + c^{2} + 1}$ $= \frac{3 a^{2} b^{2} c^{2} + 2 (a^{2} b^{2} + b^{2} c^{2} + a^{2} c^{2}) + a^{2} + b^{2} + c^{2}}{a^{2} b^{2} c^{2} + a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2} + a^{2} + b^{2} + c^{2} + 1}$ $= \frac{3 a^{2} b^{2} c^{2} + 2 (a^{2} b^{2} + b^{2} c^{2} + a^{2} c^{2}) + a^{2} + b^{2} + c^{2}}{3 a^{2} b^{2} c^{2} + 2 (a^{2} b^{2} + b^{2} c^{2} + a^{2} c^{2}) + a^{2} + b^{2} + c^{2}}$ $(p u t t i n g t h e v a l u e o f 1 i n d e n o m i n a t o r$ $a^{2} b^{2} + b^{2} c^{2} + a^{2} c^{2} + 2 a^{2} b^{2} c^{2} = 1)$ $= 1$
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