CosecA-1-cosecA-1-cosA-1-sinA 2

CosecA 1/cosecA+1 In LHS, multiply the numerator and denominator with sin A. =(1 sinA)/(1+sinA) Again multiply the numerator and denominator with (1+sinA). =(1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

cosecA-1/cose...

Question

$\frac{c o s e c A - 1}{c o s e c A + 1} = {(\frac{c o s A}{1 + s i n A})}^{2}$

Open in App

Solution

$\frac{c o s e c A - 1}{c o s e c A + 1}$

In LHS, multiply the numerator and denominator with sin A.

= $\frac{(1 - s i n A)}{(1 + s i n A)}$

Again multiply the numerator and denominator with (1+sinA).

= $\frac{(1 - s i n A)}{(1 + s i n A)}$ $\times$ $\frac{(1 + s i n A)}{(1 + s i n A)}$

= $\frac{1 - s i n^{2} A}{(1 + s i n A)^{2}}$

= $\frac{c o s^{2} A}{(1 + s i n A)^{2}}$

= ( $\frac{c o s A}{(1 + s i n A)})^{2}$

Since, LHS = RHS, it is proved.

Suggest Corrections

Similar questions

Q. Prove that

\frac{cot A - cos A}{cot A + cos A} = \frac{cos e c A - 1}{cos e c A + 1}

prove that: cotA-cosA/ cotA +cosA = cosecA-1/cosecA+1

Prove: $\frac{cot A + c o s e c A - 1}{cot A - c o s e c A + 1} = \frac{1 + cos A}{sin A}$

Prove the following trigonometric identities:

$\frac{c o t A - c o s A}{c o t A + c o s A} = \frac{c o s e c A - 1}{c o s e c A + 1}$

Prove:

\frac{c o t A + c o s e c A - 1}{c o t A - c o s e c A + 1} = \frac{1 + c o s A}{s i n A} = c o s e c θ + c o t θ = \frac{s i n A}{1 - c o s A}

Join BYJU'S Learning Program

cosecA−1cosecA+1=(cosA1+sinA)2

cosecA−1cosecA+1 In LHS, multiply the numerator and denominator with sin A. =(1−sinA)(1+sinA) Again multiply the numerator and denominator with (1+sinA). =(1−sinA)(1+sinA) × (1+sinA)(1+sinA) =1−sin2A(1+sinA)2 =cos2A(1+sinA)2 = (cosA(1+sinA))2 Since, LHS = RHS, it is proved.

$\frac{c o s e c A - 1}{c o s e c A + 1} = {(\frac{c o s A}{1 + s i n A})}^{2}$