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Question Number 82886 by 1406 last updated on 25/Feb/20

$$prove$$$${(tanx+{cot}^{2}x)}^2={sex}^{2}x+{cosec}^{2}x$$

Commented by jagoll last updated on 25/Feb/20

$$\Rightarrow \mathrm{Rhs}:\sec {}^2\mathrm{x}+\mathrm{cosec}{}^2\mathrm{x}=$$$$\frac{1}{\cos {}^2\mathrm{x}}+\frac{1}{\sin {}^2\mathrm{x}}=\frac{4}{{\left(2\sin \mathrm{xcos}\mathrm{x}\right)}^2}$$$$=\frac{4}{\sin {}^{2}2\mathrm{x}}$$$$\Rightarrow \mathrm{Lhs}:{(\frac{\sin \mathrm{x}}{\cos \mathrm{x}}+\frac{\cos {}^2\mathrm{x}}{\sin {}^2\mathrm{x}})}^2=$$$${\left(\frac{\sin {}^3\mathrm{x}+\cos {}^3\mathrm{x}}{\frac{1}{2}\sin \mathrm{xsin}2\mathrm{x}}\right)}^2=4\left(\frac{{(\sin {}^3\mathrm{x}+\cos {}^3\mathrm{x})}^2}{\sin {}^2\mathrm{x}\sin {}^{2}2\mathrm{x}}\right)$$$$\frac{4\left[{(\sin \mathrm{x}+\cos \mathrm{x})}^2{(1-\frac{1}{2}\sin 2\mathrm{x})}^2\right]}{\sin {}^2\mathrm{x}\sin {}^{2}2\mathrm{x}}$$$$=\frac{2(1+\sin 2\mathrm{x})(2-\sin 2\mathrm{x})}{\sin \mathrm{x}\sin {}^{2}2\mathrm{x}}$$$$\mathrm{question}\mathrm{wrong}$$

Commented by MJS last updated on 25/Feb/20

$$\mathrm{it}\mathrm{must}\mathrm{be}$$$${(tanx+cotx)}^2={sex}^{2}x+{cosec}^{2}x$$

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