Question

prove the following identity (sin theta+sec theta)²(cos theta+cosec theta)²=1+(sec theta+cosec theta)²​

Answer 1

Correct Question :-

To Prove :-

$\sf (sin \theta+sec \theta)^2 (cos\theta+cosec\theta)^2 = 1 +(sec \theta cosec \theta )^2$

Solution :-

$\sf L.H.S = (sin \theta+sec \theta)^2 (cos\theta +cosec\theta)^2$

$\bullet \bf \ sec \theta = \dfrac{1}{cos\theta} \\\\\bullet \ cosec\theta = \dfrac{1}{sin \theta}$

         $= \sf \left( sin \theta + \dfrac{1}{cos \theta} \right)^2 \left( cos \theta+ \dfrac{1}{sin\theta} \right)^2 \\\\= \left( \dfrac{sin\theta cos\theta +1 }{cos\theta} \right)^2 \left( \dfrac{sin\theta cos \theta+1}{sin\theta} \right)^2 \\\\= \dfrac{(sin \theta cos \theta+1)^2}{cos ^2 \theta} \times \dfrac{(sin \theta cos\theta +1)^2}{sin^2 \theta} \\\\= (sin \theta cos+1)^2 \times \left[\dfrac{1}{cos ^2\theta}+\dfrac{1}{sin^2 \theta }\right]$

         $=\sf (sin \theta cos\theta+1)^2 \times \left[ \dfrac{sin^2\theta +cos^2\theta} {sin^2 \theta cos^2 \theta} \right]$

$\bullet \bf \ sin^2\theta +cos^2\theta = 1$

         $= \sf (sin \theta cos \theta +1)^2\times \left[ \dfrac{1}{sin^2\theta cos^2\theta} \right]\\\\= (sin\theta cos\theta +1)^2 \times \left[ \dfrac{1}{sin \theta cos\theta} \right]^2\\\\=\left( \dfrac{sin\theta cos\theta+1}{sin\theta cos\theta } \right)^2 \\\\= \left( \dfrac{sin\theta cos\theta} {sin\theta cos \theta} +\dfrac{1}{sin\theta} \times \dfrac{1}{cos\theta} \right)^2 \\\\= 1 + ( cosec \theta sec\theta )^2\\\\= 1+ ( sec\theta cosec\theta )^2 \\\\= R.H.S$

Hence proved.

Answer 2

Step-by-step explanation:

$Solution :-\sf L.H.S = (sin \theta+sec \theta)^2 (cos\theta +cosec\theta)^2L.H.S=(sinθ+secθ)2(cosθ+cosecθ)2\begin{gathered}\bullet \bf \ sec \theta = \dfrac{1}{cos\theta} \\\\\bullet \ cosec\theta = \dfrac{1}{sin \theta}\end{gathered}∙ secθ=cosθ1∙ cosecθ=     \begin{gathered}= \sf \left( sin \theta + \dfrac{1}{cos \theta} \right)^2 \left( cos \theta+ \dfrac{1}{sin\theta} \right)^2 \\\\= \left( \dfrac{sin\theta cos\theta +1 }{cos\theta} \right)^2 \left( \dfrac{sin\theta cos \theta+1}{sin\theta} \right)^2 \\\\= \dfrac{(sin \theta cos \theta+1)^2}{cos ^2 \theta} \times \dfrac{(sin \theta cos\theta +1)^2}{sin^2 \theta} \\\\= (sin \theta cos+1)^2 \times \left[\dfrac{1}{cos ^2\theta}+\dfrac{1}{sin^2 \theta }\right]\end{gathered=(sinθ+cosθ1)2(cosθ+sinθ1)2=(cosθsinθcosθ+1)2(sinθsinθcosθ+1)2=cos2θ(sinθcosθ+1)2×sin2θ(sinθcosθ+1)2=(sinθcos+1)2×[cos2θ1+sin2θ1]l=\sf (sin \theta cos\theta+1)^2 \times \left[ \dfrac{sin^2\theta +cos^2\theta} {sin^2 \theta cos^2 \theta} \right]=(sinθcosθ+1)2×[sin2θcos2θsin2θ+cos2θ\bullet \bf \ sin^2\theta +cos^2\theta = 1∙ sin2θ+cos2θ=1\begin{gathered}= \sf (sin \theta cos \theta +1)^2\times \left[ \dfrac{1}{sin^2\theta cos^2\theta} \right]\\\\= (sin\theta cos\theta +1)^2 \times \left[ \dfrac{1}{sin \theta cos\theta} \right]^2\\\\=\left( \dfrac{sin\theta cos\theta+1}{sin\theta cos\theta } \right)^2 \\\\= \left( \dfrac{sin\theta cos\theta} {sin\theta cos \theta} +\dfrac{1}{sin\theta} \times \dfrac{1}{cos\theta} \right)^2 \\\\= 1 + ( cosec \theta sec\theta )^2\\\\= 1+ ( sec\theta cosec\theta )^2 \\\\= R.H.S\end{gathered}=(sinθcosθ+1)2×[sin2θcos2θ1]=(sinθcosθ+1)2×[sinθcosθ1]2=(sinθcosθsinθcosθ+1)2=(sinθcosθsinθcosθ+sinθ1×cosθ1)2=1+(cosecθsecθ)2=1+(secθcosecθ)2=R.H.S</p><p></p><p>Hence proved.</p><p></p><p>$ Thank you .

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