Question

Prove that

$\frac{tan(θ) }{1 - \ \cot(θ) ) } + \frac{ \ \cot(θ) ) }{1 - \tan(θ) } \\ \\ = 1 + \sec(θ)cosecθ$
​

Answer 1

FORMULA USED:-

$\underline{\boxed{\sf a³-b³ = (a-b)(a²+ab+b²)}}$

L.H.S :-

$\sf \green{ \dfrac{ \tan( \theta) }{1 - \cot(\theta ) } + \dfrac{ \cot(\theta) }{1 - \tan(\theta) } }$

$\bigstar{ \pmb {\red{{ \sf Required \: solution }}}}$

$\implies \sf \pink{ \dfrac{ \tan(\theta) }{1 - \dfrac{1}{ \tan(\theta) } } + \dfrac{\dfrac{1}{ \tan(\theta) }}{1 - \tan( \theta) } }$

$\implies\sf \purple{ \dfrac{ \tan(\theta) }{ \dfrac{ \tan(\theta) - 1 }{ \tan(\theta) } } + \dfrac{ \dfrac{1}{ \tan(\theta) } }{1 - \tan(\theta) } }$

$\implies \sf \pink{ \dfrac{ \tan ^{2}\theta }{ \tan\theta - 1} + \dfrac{1}{ \tan\theta ( \tan\theta - 1)}}$

$\implies\sf \purple{ \dfrac{ \tan ^{2} \theta }{ \tan\theta - 1 } - \dfrac{1}{ \tan\theta( \tan\theta - 1 )}}$

$\implies\sf \pink{ \dfrac{ \tan ^{3} \theta - 1 }{ \tan\theta( \tan\theta - 1 ) } }$

$\implies \sf \purple{ \dfrac{ \cancel{(\tan\theta - 1}) ( { \tan}^{2}\theta + 1 + \tan\theta )}{ \tan\theta( \cancel{ \tan\theta - 1 )}} }$

$\implies\sf \pink{ \dfrac{ \tan ^{2} \theta + 1 + \tan\theta}{ \tan\theta} }$

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

Now, Convert this value in terms of sin and cos

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

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

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

$\implies \sf \blue{ \sec\theta \csc\theta + 1 }$

Not appreciated for de answer :(

Answer 2

Answer:

ninte ans arum report cheyyathe dlt ayath kando...

alle ingane kanikkum.