Question

prove that: tan A / ( 1 - cot A )+ cot A / ( 1 - tan A ) =1+ sec A .cosec A.

Answer 1

$\underline{\underline{\Huge{\textbf{Question:}}}}$

$Prove\: that \: \: \dfrac{tan\: A}{1-cot\: A}+\dfrac{cot\: A}{1-tan\: A}= 1+sec\: A\: cosec\: A$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\underline{\Large{\textsf{Step-1:}}}$
Consider LHS

$\underline{\textbf{LHS\: =}}$

$\implies \dfrac{tan\: A}{1-cot\: A}+\dfrac{cot\: A}{1-tan\: A}$



$\underline{\Large{\textsf{Step-2:}}}$
Express $cot\: A$ in terms of $tan\: A$ and simplify

We have,
$\bigstar \: \mathbf{cot\: \theta = \dfrac{1}{tan\: \theta}}$

$\implies \dfrac{tan\: A}{1-\frac{1}{tan\: A}}+\dfrac{\frac{1}{tan\: A}}{1-tan\: A}$

$\implies \dfrac{tan\: A}{\frac{tan\: A-1}{tan\: A}}+\dfrac{1}{tan\: A(1-tan\: A)}$

$\implies \dfrac{tan^{2}\: A}{tan\: A-1}+\dfrac{1}{tan\: A(1-tan\: A)}$



$\underline{\Large{\textsf{Step-3:}}}$
Express the two terms with same denominator

$\implies \dfrac{-tan^{2}\: A}{1-tan\: A}+\dfrac{1}{tan\: A(1-tan\: A)}$



$\underline{\Large{\textsf{Step-4:}}}$
Express the two terms as a single term

$\implies \dfrac{-tan^{3}\: A+1}{tan\: A(1-tan\: A)}$



$\underline{\Large{\textsf{Step-5:}}}$
Using the Identity
$\bigstar \: \: \mathbf{a^{3}-b^{3}= (a-b)(a^{2}+ab+b^{2}}$

$\implies \dfrac{1^{3}-tan^{3}\: A}{tan\: A(1-tan\: A)}$

$\implies \dfrac{(1 -tan\: A)(1+tan^{2}\: A+tan\: A)}{tan\: A(1-tan\: A)}$



$\underline{\Large{\textsf{Step-6:}}}$
Cancel out the common term in Numerator and Denominator

$\implies \dfrac{\cancel{(1 -tan\: A)}(1+tan^{2}\: A+tan\: A)}{tan\: A\cancel{(1-tan\: A)}}$

$\implies \dfrac{(1+tan^{2}\: A+tan\: A)}{tan\: A}$



$\underline{\Large{\textsf{Step-7:}}}$
Using the Identity
$\bigstar\: \mathbf{sec^{2}\: \theta-tan^{2}\: \theta=1}$

$\implies sec^{2}\: A= (1+tan^{2}\: A)$

$\implies \dfrac{sec^{2}\: A+tan\: A}{tan\: A}$



$\underline{\Large{\textsf{Step-8:}}}$
Split the fraction into two terms

$\implies \dfrac{sec^{2}\: A}{tan\: A}+\dfrac{\cancel{tan\: A}}{\cancel{tan\: A}}$



$\underline{\Large{\textsf{Step-9:}}}$
Using the Identity
$\bigstar \: \mathbf{cot\: \theta = \dfrac{1}{tan\: \theta}}$

$\implies (sec^{2}\: A \dfrac{1}{tan\: A})+1$

$\implies (sec^{2}\: A\: cot\: A)+1$



$\underline{\Large{\textsf{Step-10:}}}$
Express $sec\: A$ and $cot\: A$ in terms of $cos\: A$

We have,
$\bigstar \: \mathbf{sec\: \theta= \dfrac{1}{cos\: \theta}}$

$\bigstar\: \mathbf{cot\: \theta= \dfrac{cos\: \theta}{sin\: \theta}}$

$\implies \left(\dfrac{1}{cos^{2}\: A}\times \: \dfrac{cos\: A}{sin\: A}\right)+1$



$\underline{\Large{\textsf{Step-11:}}}$
cancel out the common term in both numerator and denominator

$\implies \left(\dfrac{1}{cos^{\cancel{2}}\: A}\times \: \dfrac{\cancel{cos\: A}}{sin\: A}\right)+1$

$\implies \left(\dfrac{1}{cos\: A} \times \dfrac{1}{sin\: A}\right)+1$



$\underline{\Large{\textsf{Step-12:}}}$
Using the Identity
$\bigstar \: \mathbf{sec\: \theta= \dfrac{1}{cos\: \theta}}$

$\bigstar\: \mathbf{cosec\: \theta= \dfrac{1}{sin\: \theta}}$

$\implies \left(sec\: A \times Cosec\: A\right)+1$

$\implies 1+sec\: A\: Cosec\: A$

$\underline{\textbf{= RHS\:}}$