Question

(Cosec A- Sin A) (Sec A - Cos A) = 1/ tan A + cot A​

Answer 1

Given :-

$\bullet \sf\ \ (CosecA-SinA)(SinA-CosA)=\dfrac{1}{tanA+cotA}$

Solution :-

• Taking LHS -

$\bullet\sf\ CosecA= \dfrac{1}{SinA}\ \ ;\ \bullet\sf\ \ SecA=\dfrac{1}{CosA}\\ \\ \\ :\implies\sf\ \bigg\{\dfrac{1}{SinA}-SinA\bigg\}\bigg\{\dfrac{1}{CosA}-CosA\bigg\}\\ \\ \\ :\implies\sf\ \bigg\{\dfrac{1-Sin^2A}{SinA}\bigg\}\bigg\{\dfrac{1-Cos^2A}{CosA}\bigg\}\\ \\ \\ \therefore\underline{\sf\ Sin^2\theta+Cos^2\theta=1}\\ \\ \\ :\implies\sf\ \bigg\{\dfrac{\cancel{Cos^2A}}{\cancel{SinA}}\bigg\}\times \bigg\{\dfrac{\cancel{Sin^2A}}{\cancel{CosA}}\bigg\}\\ \\ \\ :\implies\sf\ \underline{\boxed{\purple{\sf\ \ CosA\ SinA}}}------eq.(i)$

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• Now RHS :-

$:\implies\sf\ \dfrac{1}{TanA+CotA}\\ \\ \\ \bullet\sf\ \ tanA=\dfrac{SinA}{CosA}\ \;\ \bullet\sf\ CotA=\dfrac{CosA}{SinA}\\ \\ \\ :\implies\sf\ \Bigg\{\dfrac{1}{\frac{SinA}{CosA}+\frac{CosA}{SinA}}\Bigg\}\\ \\ \\ :\implies\sf\ \Bigg\{\dfrac{1}{\frac{Sin^2A+Cos^2A}{SinACosA}}\Bigg\}\\ \\ \\ :\implies\sf\ 1\times \bigg\{\dfrac{SinACosA}{Sin^2A+Cos^2A}\bigg\}\\ \\ \\ :\implies\sf\ \bigg\{\dfrac{SinACosA}{1}\bigg\}\\ \\ \\ :\implies\sf\ \underline{\boxed{\red{\sf\ \ CosA\ SinA}}}------eq.(ii)$

• From equation (i) and (ii)

$\sf\ \ LHS=RHS\ \ \ \ Hence\ Proved !!$

Answer 2

Answer:

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