Question

prove that: (sin thita + sec thita ) square + ( cos thita + cosec thita) square = (1+ sec thita cosec thita ​

Answer 1

Correct Question :-

Prove that : $\sf ( sin \theta+sec \theta)^2+(cos\theta+cosec \theta )^2= (1+sec\theta cosec\theta)^2$ .

Solution :-

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

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

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

Taking  $\sf (sin\theta cos\theta +1)^2$ as a common factor.

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

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

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

Hence proved.