Question

Prove the following identities, where the angles involved are acute angles for which the

expressions are defined.

a. cosA/(1+sinA)+(1+sinA)/cosA=2secA.

b. (1+secA)/secA=sin2A/(1−cosA)

c. √(1 + sinA)/(1−sinA) =secA+tanA​

Answer 1

Answer:

1.  cosA / (1+sinA) + (1+sinA) / cosA = 2secA.

Taking the L.H.S. we have,

cosA / (1+sinA) + (1+sinA) / cosA

= cos^2A + (1 + sinA)^2 / cosA (1+sinA)

= cos^2A + 1 + sin^2A + 2sinA / cosA (1+sinA)

= cos^2A + sin^2A + 1 + 2sinA / cosA (1+sinA)

= 1 + 1 + 2sinA / cosA (1+sinA)

= 2 + 2sinA / cosA (1+sinA)

= 2 (1+sinA) / cosA (1+sinA)

= 2 / cosA

= 2secA  

= R.H.S

2. (1+secA) / secA = sin^2A / (1−cosA)

Taking the R.H.S. we have

sin^2A / (1−cosA)

1 - cos^2A / 1 - cosA

= ( 1 - cosA ) ( 1 + cosA ) / 1 - cosA

= 1 + cosA

= 1 + 1/secA

= 1 + secA / secA

= L.H.S

3. √(1 + sinA) / (1−sinA) = secA + tanA​

Taking the L.H.S. we have

= $\sqrt ( 1 + sinA ) / ( 1 - sinA )$

= $\sqrt ( 1 + sinA ) / ( 1 - sinA ) * ( 1 + sinA ) / ( 1 + sinA )$

= $\sqrt ( 1 + sinA )^2 / ( 1 - sin^2A )$

= $\sqrt ( 1 + sinA )^2 / 1 - ( 1 - cos^2A )$

= $\sqrt ( 1 + sinA )^2 / 1 - 1 + cos^2A$

= $\sqrt ( 1 + sinA )^2 / cos^2A$

= $\sqrt 1 / cos^2A + sin^2A / cos^2A$

= $\sqrt sec^2A + tan^2A$

= secA + tanA

= R.H.S

Step-by-step explanation:

Answer 2

Step-by-step explanation:

step step explanation in image