Question

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Answer 1

Answer:

$\longrightarrow\tt{\bigg(\dfrac{tan\:A}{1-cot\:A}+\dfrac{cot\:A}{1-tan\:A}\bigg)}$

Now, we know that cot A = 1/tan A.

Hence,

$\longrightarrow\sf{\bigg(\dfrac{tan\:A}{1-\dfrac{1}{tan\:A}}+\dfrac{\dfrac{1}{tan\:A}}{1-tan\:A}\bigg)}$

Simplifying,

$\longrightarrow\sf{\Bigg(\dfrac{\dfrac{tan\:A}{1}}{\dfrac{tan\:A-1}{tan\:A}}+\dfrac{\dfrac{1}{tan\:A}}{\dfrac{1-tan\:A}{1}}\Bigg)}$

$\longrightarrow\sf{\bigg(\dfrac{tan\:A}{1}\times\dfrac{tan\:A}{tan\:A-1}\bigg)+\bigg(\dfrac{1}{tan\:A}\times \dfrac{1}{1-tan\:A}\bigg)}$

$\longrightarrow\sf{\bigg(\dfrac{tan^2\:A}{tan\:A-1}+\dfrac{1}{tan\:A(1-tan\:A)}\bigg)}$

Taking out common,

$\longrightarrow\sf{\Bigg(\dfrac{1}{tan\:A-1}\bigg(\dfrac{tan^2\:A}{1}+\dfrac{1}{tan\:A}\bigg)\Bigg)}$

Taking LCM,

$\longrightarrow\sf{\Bigg(\dfrac{1}{tan\:A-1}\bigg(\dfrac{tan^3\:A-1}{tan\:A}\bigg)\Bigg)}$

Using the identity – (a³-b³) = (a-b)(a²+b²+ab).

$\longrightarrow\sf{\Bigg(\dfrac{1}{\cancel{tan\:A-1}}\bigg(\cancel{(tan\:A-1)}\times \dfrac{tan^2\:A+1+tan\:A}{tan\:A}\bigg)\Bigg)}$

Hence, we are left with,

$\longrightarrow\sf{\Bigg( \dfrac{tan^2\:A+1+tan\:A}{tan\:A}\Bigg)}$

Now, we know that, sec² A = tan² A + 1.

Hence,

$\longrightarrow\sf{\Bigg( \dfrac{sec^2\:A+tan\:A}{tan\:A}\Bigg)}$

Now simplifying,

$\longrightarrow\sf{\Bigg( \dfrac{sec^2}{tan\:A}+\cancel{\dfrac{tan\:A}{tan\:A}}\Bigg)}$

Now, we are left with,

$\longrightarrow\sf{\Bigg( \dfrac{sec^2}{tan\:A}+1\Bigg)}$

Simplifying to get final answer,

$\longrightarrow\sf{\Bigg( \dfrac{\dfrac{1}{cos^2\:A}}{\dfrac{sin\:A}{cos\:A}}+1\Bigg)}$

$\longrightarrow\sf{\Bigg( \dfrac{\dfrac{1}{cos\:A}}{\dfrac{sin\:A}{1}}+1\Bigg)}$

$\longrightarrow\sf{\Bigg( \dfrac{1}{cos\:A}\times \dfrac{1}{sin\:A}+1\Bigg)}$

Hence, the final answer is :-

$\longrightarrow{\boxed{\pink{\bf{sec\:A\:cosec\:A+1}}}}$

Answer 2

Step-by-step explanation:

_________________________

$\bf \huge Given \: \:$

$\bf tanA/(1 - cotA) + cotA/(1 - tanA)\: \:$

$\bold{\implies \frac{sinA/cosA}{1 - cosA/sinA} + \frac{cosA/sinA}{1 - sinA/cosA}}</p><p>$

$\bold{\implies \frac{sin^2A}{cosA(sinA - cosA)} + \frac{cos^2A}{sinA(cosA - sinA)}}$

$\bold{\implies \frac{sin^2A}{cosA(sinA - cosA)} - \frac{cos^2A}{sinA(sinA - cosA)}}</p><p>$

$\bold{\implies \frac{sin^3A - cos^3A}{sinA.cosA(sinA - cos)}}</p><p>$

$\bold{\implies \frac{\cancel{(sinA - cosA)}(sin^2A + cos^2A + sinAcosA)}{sinA.cosA\cancel{(sinA - cosA)}}}</p><p>$

$\bf[as, a^3 - b^3 = (a - b)(a^2+ b^2+ ab) \: \:$

$\bf (sin^2+ cos^2 = 1]\: \:$

Applying the Formula

$\bold{\implies \frac{(1 + sinA.cosA)}{sinA.cosA}}$

$\bold{\implies \frac{1}{sinA.cosA} + \frac{sinA.cosA}{sinA.cosA}}</p><p>$

$\bold{\implies secA.cosecA + 1}$

$\bold{\implies \frac{tanA}{(1 - 1/tanA)} + \frac{(1/tanA)}{(1 - tanA)}} </p><p>$

$\bold{\implies \frac{tan^2A}{(tanA - 1)} + \frac{1}{tanA(1 - tanA)}}</p><p> \\ \\ </p><p></p><p>\bold{\implies \frac{1}{tan(1 - tanA)} - \frac{tan^2A}{(1 - tanA)}}$

$\bold{\implies \frac{1 - tan^3A}{tanA(1 - tanA)}}$

$\bold{\implies \frac{\cancel{(1 - tanA)}\{(1 + tan^2A) + tanA\}}{tanA\cancel{(1 - tanA)}}}$

$\bold{\implies \frac{sec^2A + tanA}{tanA}}$

$\bold{\implies \frac{1}{cos^2A}{cosA}{sinA} + \frac{tanA}{tanA}}$

$\bold\red{{\implies secA.cosecA + 1}}$

option a is Right answer