Question

prove that (sec A + tanA)(1 - sinA)= cosA​

Answer 1

Answer:

To prove : (\sec A+\tan A)(1-\sin A)=\cos A(secA+tanA)(1−sinA)=cosA

Proof :

Taking LHS,

LHS=(\sec A+\tan A)(1-\sin A)LHS=(secA+tanA)(1−sinA)

Write, \sec A=\frac{1}{\cos A}\ , \tan A=\frac{\sin A}{\cos A}secA=

cosA

1

,tanA=

cosA

sinA

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

cosA

1

+

cosA

sinA

)(1−sinA)

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

cosA

1+sinA

)(1−sinA)

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

cosA

(1+sinA)(1−sinA)

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

cosA

1

2

−sin

2

A

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

cosA

1−sin

2

A

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

cosA

cos

2

A

LHS=\cos ALHS=cosA

LHS=RHSLHS=RHS

Hence proved.