Question

Sin^2θtanθ+cos^2θcotθ+2sinθcosθ=tanθ+cotθ
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Answer 1

`\text{L.H.S}\\\\=\sin^2 \theta \tan \theta + \cos^2 \theta \cot \theta+2\sin \theta \cos \theta \\\\\\=\sin^2 \theta \cdot (\sin \theta)/(\cos \theta) + \cos^2 \theta \cdot (\cos \theta)/(\sin \theta) + 2\sin \theta \cos \theta\\\\\\=(\sin^3 \theta)/(\cos \theta)+ (\cos^3 \theta)/(\sin \theta)+ 2\sin \theta \cos \theta\\\\\\=(\sin^4 \theta + \cos^4 \theta)/(\cos \theta \cdot \sin \theta)+ 2\sin \theta \cos \theta\\\\`

`=((\sin^2 \theta)^2 + (\cos^2 \theta)^2 )/(\cos \theta \sin \theta)+ 2\sin \theta \cos \theta\\\\\\=((\sin^2 \theta + \cos^2 \theta)^2-2\sin^2 \theta \cos^2 \theta)/(\cos \theta \sin \theta)+ 2\sin \theta \cos \theta\\\\\\=(1-2 \sin^2 \theta \cos^2 \theta)/(\cos \theta \sin \theta)+ 2\ sin \theta \cos \theta\\\\\\=(1-2\sin^2 \theta \cos^2 \theta+2\sin^2 \theta \cos^2 \theta)/(\cos \theta \sin \theta)\\\\=\frac 1{\cos \theta \sin \theta}\\\\`

`=(\sin^2 \theta + \cos^2 \theta)/(\sin \theta \cos \theta)\\\\\\=(\sin^2 \theta)/(\cos \theta \sin \theta)+(\cos^2 \theta)/(\cos \theta \sin \theta)\\\\\\=(\sin \theta)/(\cos \theta)+(\cos \theta)/(\sin \theta)\\\\\\=\tan \theta + \cot \theta\\\\=\text{R.H.S}\\\\\text{Proved.}`