Question

prove that 1+secA /sec A = sin ^2 A / 1- cos A​

Answer 1

Answer:

Step-by-step explanation:

rove that 1+secA /sec A = sin ^2 A / 1- cos A​

Answer 2

Step-by-step explanation:

Taking LHS,

LHS=\frac{1+\sec A}{\sec A}LHS=

secA

1+secA

Substitute, \sec A=\frac{1}{\cos A}secA=

cosA

1

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

cosA

1

1+

cosA

1

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

cosA

1

cosA

cosA+1

LHS=\frac{\cos A+1}{\cos A}\times \cos ALHS=

cosA

cosA+1

×cosA

LHS=\cos A+1LHS=cosA+1

Multiply and divide by \cos A-1cosA−1

LHS=(\cos A+1)\times \frac{\cos A-1}{\cos A-1}LHS=(cosA+1)×

cosA−1

cosA−1

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

cosA−1

(cosA+1)(cosA−1)

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

cosA−1

cos

2

A−1

2

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

cosA−1

−sin

2

A

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

1−cosA

sin

2

A

LHS=RHSLHS=RHS

Hence proved.