Question

if cos ( A + B ) = 0 and cot ( A-B ) = √3 find the value of secA.tanB - cotA . sinB

Answer 1

Answer:

$\sec A\tan B-\cot A\sin B=\frac{\sqrt3}{2}$

Step-by-step explanation:

Given : $\text{If } \cos(A+B)= 0 \text{ and } \cot(A-B)=\sqrt{3}$

To find : The value of $\sec A\tan B-\cot A\sin B$

Solution :  

We know,  

1)  cos(A+B)=0

⇒cos(A+B) = cos 90°

⇒ A + B = 90°....(i)

2)  cot(A - B) = √3

⇒ Cot(A - B) = cot 30°

⇒ A - B = 30°  ........(ii)

On adding (i) & (ii)

2A = 120

A = 60°

Put A = 60° in (i)

A + B = 90

60 + B = 90

B = 30

Substitute the value of A and B in the value we get,

$\sec A\tan B-\cot A\sin B$

$=\sec (60)\tan (30)-\cot (60)\sin (30)$ 

$=(2)(\frac{1}{\sqrt3})-(\frac{1}{\sqrt3})(\frac{1}{2})$   

$=\frac{2}{\sqrt3}-\frac{1}{2\sqrt3}$

$=\frac{4-1}{2\sqrt3}$

$=\frac{3}{2\sqrt3}$

$=\frac{\sqrt3}{2}$

Therefore, $\sec A\tan B-\cot A\sin B=\frac{\sqrt3}{2}$