Question

sinA×tanA/1-cosA=1+secA prove by taking LHS​

Answer 1

Answer:

Just mind my handwriting

Step-by-step explanation:

Hope it helped

Answer 2

Answer:

Answer and explanation:

To prove : \frac{\sin A\cdot\tan A}{1-\cos A}=1+\sec A1−cosAsinA⋅tanA=1+secA

Proof :

Taking LHS,

LHS=\frac{\sin A\cdot\tan A}{1-\cos A}LHS=1−cosAsinA⋅tanA

LHS=\frac{\sin A\cdot\frac{\sin A}{\cos A}}{1-\cos A}LHS=1−cosAsinA⋅cosAsinA

LHS=\frac{\frac{\sin^2 A}{\cos A}}{1-\cos A}LHS=1−cosAcosAsin2A

LHS=\frac{\frac{1-\cos^2 A}{\cos A}}{1-\cos A}LHS=1−cosAcosA1−cos2A

LHS=\frac{\frac{(1-\cos A)(1+\cos A)}{\cos A}}{1-\cos A}LHS=1−cosAcosA(1−cosA)(1+cosA)

LHS=\frac{1+\cos A}{\cos A}LHS=cosA1+cosA

LHS=\sec A+1LHS=secA+1

LHS=RHSLHS=RHS