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Question Number 218889 by Spillover last updated on 17/Apr/25

	Answered by mehdee7396 last updated on 17/Apr/25

	$t a n x + c o t x = - 1$ ${t a n}^{3} + {c o t}^{3} x = {(t a n x + c o t x)}^{3} - 3 (t a n x \times c o t x) (t a n x + c o t x)$ $= - 1 + 3 = 2 = 2 m^{2} \Rightarrow m = \pm 1$
	Answered by som(math1967) last updated on 17/Apr/25

	$(s e c x + c o s e c x) (s i n x + c o s x) = 1$ $\Rightarrow \frac{{(\sin x + \cos x)}^{2}}{s i n x c o s x} = 1$ $\Rightarrow 1 + 2 s i n x c o s x = s i n x c o s x$ $\Rightarrow s i n x c o s x = - 1$ ${t a n}^{3} x + {c o t}^{3} x = 2 m^{2}$ $\Rightarrow \frac{{s i n}^{6} x + {c o s}^{6} x}{{s i n}^{3} {x c o s}^{3} x} = 2 m^{2}$ $\Rightarrow \frac{{(\sin^{2} x + \cos^{2} x)}^{3} - 3 {s i n}^{2} x \cos^{2} x ({s i n}^{2} x + {c o s}^{2} x)}{{(- 1)}^{3}}$ $= 2 m^{2}$ $\Rightarrow - 1 + 3 = 2 m^{2}$ $m = \pm 1$
	Answered by Spillover last updated on 17/Apr/25

	Answered by Spillover last updated on 17/Apr/25

	Answered by Spillover last updated on 17/Apr/25

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