Question

Trigonometry - Verify Identity:


cos (x - y) / sin (x + y) = 1 + cotxcoty / cotx + coty

Answer 1

Answer:

The proof is given below.

Step-by-step explanation:

Given:

The identity to verify is given as:

$\frac{\cos (x-y)}{\sin (x+y)}=\frac{1+\cot x.\cot y}{\cot x+\ cot y}$

Consider the left hand side of the identity.

$\because \cos (A-B)=\cos A\cdot \cos B+\sin A\cdot \sin B\\ \sin (A+B)=\sin A\cdot \cos B+\sin B\cdot \cos A$

$= \frac{\cos (x-y)}{\sin (x+y)}\\=\frac{\cos x\cdot \cos y+\sin x\cdot \sin y}{\sin x\cdot \cos y+\sin y\cdot \cos x}$

Dividing both numerator and denominator by $\sin x\cdot \sin y$. This gives,

$=\frac{\frac{\cos x\cdot \cos y}{\sin x\cdot \sin y}+\frac{\sin x\cdot \sin y}{\sin x\cdot \sin y}}{\frac{\sin x\cdot \cos y}{\sin x\cdot \sin y}+\frac{\sin y\cdot \cos x}{\sin x\cdot \sin y}} \\\\\\=\frac{\cot x\cdot \cot y+1}{\cot x+\cot y}\\\\=\frac{1+\cot x.\cot y}{\cot x+\ cot y}=RHS$

We have used the identity $\cot A = \frac{\cos A}{\sin A}$ above.

Therefore,

$\frac{\cos (x-y)}{\sin (x+y)}=\frac{1+\cot x.\cot y}{\cot x+\ cot y}$

Therefore, LHS = RHS and hence proved.