Question

Sin theta(1+tan theta) +cos(1+cosec theta) =sec theta+cosec theta

Answer 1

there is fault in this question .

correct question :- Sin theta(1+tan theta) +cos(1+cot theta) =sec theta+cosec theta

basic concept used :-

• firstly we have to convert LHS in the form of sinθ and cosθ .

$\implies \: \sf \: \tan \theta \: = \dfrac{ \sin \theta}{ \cos \theta } \\ \\ \implies \: \sf \: \cot \theta\: = \frac{ \cos \theta}{ \sin \theta} \\ \\ \implies \: \sf \: \frac{1}{ \sin\theta} = \cosec \theta \\ \\ \implies \: \sf \: \frac{1}{ \cos\theta} = \sec \theta \\ \\\implies \: \sf \: { \sin}^{2} \theta \: + {cos}^{2} \theta \: = 1$

Solution :-

L.H.S

$\sf \: sin \theta \: (1 + tan \theta) + cos\theta(1 + cot\theta) \\ \\ \sf \: sin\theta \bigg ( 1 + \dfrac{ \sin\theta}{cos\theta} \bigg) + cos\theta\bigg (1 + \dfrac{cos \theta}{sin \theta} \\ \\ \sf \: sin \theta\bigg( \dfrac{cos \theta + sin \theta}{cos \theta} \bigg) + cos \theta \bigg( \dfrac{sin \theta + cos \theta}{sin \theta} \bigg) \\ \\ \sf \dfrac{sin \theta.cos \theta + {sin}^{2} \theta}{cos \theta} + \dfrac{sin \theta.cos \theta + {cos}^{2} \theta}{sin \theta} \\ \\ \sf \: \dfrac{sin \theta \big(sin\theta.cos\theta + {sin}^{2} \theta \big) + cos\theta \big(sin \theta.cos\theta + {cos}^{2} \theta }{cos\theta.sin\theta} \\ \\ \sf \: \dfrac{ {sin}^{2}\theta.cos \theta + {sin}^{3}\theta + sin \theta. {cos}^{2}\theta + {cos}^{3} \theta }{cos\theta.sin\theta} \\ \\ \sf \: \dfrac{ {sin}^{2}\theta \big(cos\theta + sin\theta \big) + {cos}^{2} \theta \big(sin\theta + cos\theta \big) }{cos\theta.sin\theta} \\ \\ \sf \: \dfrac{ \big(cos\theta + sin\theta \big) \big( {sin}^{2}\theta + {cos}^{2} \theta \big) }{cos\theta.sin\theta} \\ \\ \sf \dfrac{cos\theta + sin\theta}{cos\theta.sin\theta} \\ \\ \sf \: \dfrac{cos\theta}{ cos\theta.sin\theta} + \frac{sin \theta}{cos\theta.sin\theta} \\ \\ \sf \: \dfrac{ \cancel {cos\theta}}{ \cancel{cos\theta}.sin\theta} + \dfrac{ \cancel{sin \theta}}{cos\theta. \cancel{sin\theta} } \\ \\ \sf \: \dfrac{1}{sin \theta} + \dfrac{1}{cos\theta} \\ \\ \sf \: cosec\theta \: + sec\theta$

= R.H.S

L.H.S = R.HS

hence proved ✓✓