Question

1+sinA÷cosA=1+sin A+cosA÷1-cosA-sinA​

Answer 1

Answer:

PROOF:

Taking L.H.S.

\rm{\dashrightarrow \dfrac{(1-sinA)}{cosA} }⇢

cosA

(1−sinA)

Rationalising the denominator

\rm{\dashrightarrow \dfrac{(1-sinA)cos A}{cosA \times cos A} }⇢

cosA×cosA

(1−sinA)cosA

\rm{\dashrightarrow \dfrac{(1-sinA)cos A}{cos^2 A} }⇢

cos

2

A

(1−sinA)cosA

Using Identity:-

sin²A + cos²A = 1

cos²A = 1 – sin²A

\rm{\dashrightarrow \dfrac{(1-sinA)cos A}{1-sin^2 A} }⇢

1−sin

2

A

(1−sinA)cosA

Using Identity:

a² – b² = (a + b)(a – b)

\rm{\dashrightarrow \dfrac{(1-sinA)cos A}{(1+sin A)(1-sin A)} }⇢

(1+sinA)(1−sinA)

(1−sinA)cosA

\rm{\dashrightarrow \dfrac{\cancel{(1-sinA)}cos A}{(1+sin A)\cancel{(1-sin A)}} }⇢

(1+sinA)

(1−sinA)

(1−sinA)

cosA

\bf{\dashrightarrow \dfrac{cos A}{1 + sin A} = R.H.S. }⇢

1+sinA

cosA

=R.H.S.

\bf{\underline{\underline{ Hence \: proved}}}

Henceproved