If cot x−tan x=4 then cot^2 +(2/(sin 2x)) −tan^2 x =?


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Question Number 193533 by horsebrand11 last updated on 16/Jun/23

$If \cot x - \tan x = 4 then$ $\cot^{2} + \frac{2}{\sin 2 x} - \tan^{2} x = ?$
	Answered by MM42 last updated on 16/Jun/23

	$1 - {t a n}^{2} x = 4 t a n x \Rightarrow {t a n}^{2} x + 4 t a n x - 1 = 0$ $t a n x = - 2 \pm \sqrt{5}$ $A = \frac{1}{{t a n}^{2} x} + \frac{1 + {t a n}^{2} x}{t a n x} - {t a n}^{2} x$
	Answered by Rajpurohith last updated on 19/Jun/23

	$s a y a = t a n x a n d b = c o t x \Rightarrow a b = 1$ $g i v e n b - a = 4$ $\Rightarrow {(b + a)}^{2} - {(b - a)}^{2} = 4 a b = 4$ $\Rightarrow {(b + a)}^{2} - 16 = 4 \Rightarrow b + a = 2 \sqrt{5}$ $\Rightarrow 2 b = 4 + 2 \sqrt{5} \Rightarrow b = c o t x = 2 + \sqrt{5}$ $\Rightarrow a = t a n x = - 2 + \sqrt{5}$ $\Rightarrow {c o t}^{2} x - {t a n}^{2} x = (b + a) (b - a) = 2 \sqrt{5} . 4 = 8 \sqrt{5}$ $s o s i n 2 x = \frac{2 a}{1 + a^{2}} = \frac{2 \sqrt{5} - 4}{1 + 4 + 5 - 4 \sqrt{5}} = \frac{2 (\sqrt{5} - 2)}{10 - 4 \sqrt{5}}$ $= \frac{2 (\sqrt{5} - 2)}{2 \sqrt{5} (\sqrt{5} - 2)} = \frac{1}{\sqrt{5}}$ $\Rightarrow t h e v a l u e o f g i v e n e x p r e s s i o n i s$ ${(2 + \sqrt{5})}^{2} + 2 \sqrt{5} - {(- 2 + \sqrt{5})}^{2}$ $= 2 \sqrt{5} + (2 \sqrt{5}) (4) = 10 \sqrt{5} ◼$
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