Question

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

Answer 1

Answer:  The proof is given below.

Step-by-step explanation:  We are given to prove the following trigonometric equality :

$\dfrac{\tan A}{1\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\csc A.$

We will be using the following trigonometric formulas :

$(i)~\tan\theta=\dfrac{\sin\theta}{\cos\theta},\\\\\\(ii)~\sin^2\theta+\cos^2\theta=1.$

The proof of the given equality is as follows :

$L.H.S.\\\\\\\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}\\\\\\=\dfrac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\dfrac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}\\\\\\=\dfrac{\sin A}{\cos A}\times\dfrac{\sin A}{\sin A-\cos A}+\dfrac{\cos A}{\sin A}\times\dfrac{\cos A}{\cos A-\sin A}\\\\\\=\dfrac{\sin^2 A}{\cos A(\sin A-\cos A)}-\dfrac{\cos^2 A}{\sin A(\sin A-\cos A)}\\\\\\=\dfrac{\sin^3 A-\cos^3 A}{\sin A\cos A(\sin A-\cos A)}\\\\\\=\dfrac{(\sin A-\cos A)(\sin^2 A+\sin A\cos A+\cos^2 A)}{\sin A\cos A(\sin A-\cos A)}\\\\\\=\dfrac{\sin A\cos A+1}{\sin A\cos A}\\\\\\=1+\sec A\csc A\\\\=R.H.S.$

Hence proved.

Answer 2

Step-by-step explanation:

The proof of the given equality is as follows :

tanA1−cotA+cotA1−tanA=sinAcosA1−cosAsinA+cosAsinA1−sinAcosA  

=sin2AcosA(sinA−cosA)+cos2AsinA(cosA−sinA)

=sin3A−cos3AcosAsinA(sinA−cosA)

=sin2A+cosAsinA+cos2AcosAsinA

=1+cosAsinAcosAsinA

=1+secAcscA.

tan A/1-cot A + cot A/1-tan A = 1+sec A cosec A

HENCE PROVED.

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