Question

Prove above identity of trigonometry ​

Answer 1

We have to prove that,

$\\ \sf \implies \boxed{\dfrac{ \tan \theta }{1 - \cot \theta } + \dfrac{ \cot \theta }{1 - \tan \theta } = 1 + \tan \theta + \cot \theta}$

Solving LHS,

$\\ \sf \implies {\dfrac{ \frac{ \sin \theta}{ \cos \theta} }{1 - \frac{\cos \theta}{\sin \theta} } + \dfrac{ \frac{\cos \theta}{\sin \theta} }{1 - \frac{ \sin \theta}{ \cos \theta} } }$

$\\ \sf \implies {\dfrac{ \frac{ \sin \theta}{ \cos \theta} }{ \frac{\sin \theta - \cos \theta}{\sin \theta} } + \dfrac{ \frac{\cos \theta}{\sin \theta} }{ \frac{\cos \theta - \sin \theta}{ \cos \theta} } }$

$\\ \sf \implies \dfrac{ \sin \theta}{ \cos \theta} \times \dfrac{\sin \theta}{\sin \theta - \cos \theta} \: + \: \dfrac{\cos \theta}{\sin \theta} \times \dfrac{\cos \theta}{\cos \theta - \sin \theta}$

$\\ \sf \implies \dfrac{ \sin^{2} \theta}{ \cos \theta(\sin \theta - \cos \theta)} + \: \dfrac{\cos ^{2} \theta}{\sin \theta(\cos \theta - \sin \theta)}$

$\\ \sf \implies \dfrac{ \sin^{2} \theta}{ \cos \theta(\sin \theta - \cos \theta)} - \: \dfrac{\cos ^{2} \theta}{\sin \theta(\sin \theta - \cos \theta)}$

$\\ \sf \implies \dfrac{ \sin^{3} \theta - \cos ^{3} \theta}{ \cos \theta(\sin \theta - \cos \theta)\sin \theta}$

$\\ \sf \implies \dfrac{ \cancel{(\sin \theta - \cos \theta)}(\sin^{2} \theta + \cos^{2}\theta + \sin \theta\cos \theta)}{ \cos \theta \cancel{(\sin \theta - \cos \theta)}\sin \theta}$

$\\ \sf \implies \dfrac{ {}(\sin^{2} \theta + \cos^{2}\theta + \sin \theta\cos \theta)}{ \cos \theta {}\sin \theta}$

$\\ \sf \implies \dfrac{ 1 + \sin \theta\cos \theta}{ \cos \theta {}\sin \theta}$

$\\ \sf \implies {\dfrac{ 1}{ \cos \theta {}\sin \theta} + {\dfrac{ \sin \theta\cos \theta}{ \cos \theta {}\sin \theta} } }$

$\\ \implies \boxed{ \sf{\dfrac{ 1}{ \cos \theta {}\sin \theta} + 1 }}$

Solving RHS,

$\\ \sf \implies { 1 + \tan \theta + \cot \theta}$

$\\ \sf \implies { 1 + \dfrac{\sin \theta}{\cos \theta} + \dfrac{ \cos \theta}{ \sin \theta} }$

$\\ \sf \implies { 1 + \dfrac{\sin^{2} \theta + \cos^{2}\theta}{\cos \theta\sin \theta} }$

$\\ \sf \implies \boxed{ { 1 + \dfrac{1}{\cos \theta\sin \theta} }}$

• LHS = RHS

HENCE PROVED