Question

sinA(1+tanA)+cosA(1+cotA)=secA.cosecA​

Answer 1

QUESTION:

• Prove that: sinA(1 + tanA) + cosA(1 + cotA) = secA + cosecA

ANSWER:

To Prove:

• sinA(1 + tanA) + cosA(1 + cotA) = secA + cosecA

Proof:

We need to prove that,

$\implies\sin A(1+\tan A)+\cos A(1+\cot A)=\sec A+\csc A$

Taking LHS,

$\implies\sin A(1+\tan A)+\cos A(1+\cot A)$

We know that,

$\hookrightarrow\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

And,

$\hookrightarrow\cot\theta=\dfrac{\cos\theta}{\sin\theta}$

So,

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

Taking LCM,

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

$\implies\dfrac{\sin A(\sin A+\cos A)}{\cos A}+\dfrac{\cos A(\sin A+\cos A)}{\sin A}$

Taking LCM,

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

Taking (sin A + cos A) common,

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

We  know that,

$\hookrightarrow\sin^2\theta+\cos^2\theta=1$

So,

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

$\implies\dfrac{(\sin A+\cos A)(1)}{\sin A\cos A}$

$\implies\dfrac{\sin A+\cos A}{\sin A\cos A}$

Separating sin A + cos A,

$\implies\dfrac{\sin A}{\sin A\cos A}+\dfrac{\cos A}{\sin A\cos A}$

Cancelling like terms,

$\implies\dfrac{1}{\cos A}+\dfrac{1}{\sin A}$

We know that,

$\hookrightarrow\dfrac{1}{\cos\theta}=\sec\theta$

And,

$\hookrightarrow\dfrac{1}{\sin\theta}=\csc\theta$

So,

$\implies\dfrac{1}{\cos A}+\dfrac{1}{\sin A}$

$\implies\bf sec\,A+csc\,A$

= RHS

Hence Proved!!!