Question

If cos(a+b)=0 and cot(a-b)=√3 then find the value of the equation given in the picture​

Answer 1

Answer:

$Answer:</p><p></p><p>\sec A\tan B-\cot A\sin B=\frac{\sqrt3}{2}secAtanB−cotAsinB= </p><p>2</p><p>3</p><p> </p><p> </p><p> </p><p> </p><p></p><p>Step-by-step explanation:</p><p></p><p>Given : \text{If } \cos(A+B)= 0 \text{ and } \cot(A-B)=\sqrt{3}If cos(A+B)=0 and cot(A−B)= </p><p>3</p><p> </p><p> </p><p></p><p>To find : The value of \sec A\tan B-\cot A\sin BsecAtanB−cotAsinB</p><p></p><p>Solution : </p><p></p><p>We know, </p><p></p><p>1) cos(A+B)=0</p><p></p><p>⇒cos(A+B) = cos 90°</p><p></p><p>⇒ A + B = 90°....(i)</p><p></p><p>2) cot(A - B) = √3</p><p></p><p>⇒ Cot(A - B) = cot 30°</p><p></p><p>⇒ A - B = 30° ....(ii)</p><p></p><p>On adding (i) & (ii)</p><p></p><p>2A = 120</p><p></p><p>A = 60°</p><p></p><p>Put A = 60° in (i)</p><p></p><p>A + B = 90</p><p></p><p>60 + B = 90</p><p></p><p>B = 30</p><p></p><p>Substitute the value of A and B in the value we get,</p><p></p><p>\sec A\tan B-\cot A\sin BsecAtanB−cotAsinB</p><p></p><p>=\sec (60)\tan (30)-\cot (60)\sin (30)=sec(60)tan(30)−cot(60)sin(30) </p><p></p><p>=(2)(\frac{1}{\sqrt3})-(\frac{1}{\sqrt3})(\frac{1}{2})=(2)( </p><p>3</p><p> </p><p> </p><p>1</p><p> </p><p> )−( </p><p>3</p><p> </p><p> </p><p>1</p><p> </p><p> )( </p><p>2</p><p>1</p><p> </p><p> ) </p><p></p><p>=\frac{2}{\sqrt3}-\frac{1}{2\sqrt3}= </p><p>3</p><p> </p><p> </p><p>2</p><p> </p><p> − </p><p>2 </p><p>3</p><p> </p><p> </p><p>1</p><p> </p><p> </p><p></p><p>=\frac{4-1}{2\sqrt3}= </p><p>2 </p><p>3</p><p> </p><p> </p><p>4−1</p><p> </p><p> </p><p></p><p>=\frac{3}{2\sqrt3}= </p><p>2 </p><p>3</p><p> </p><p> </p><p>3</p><p> </p><p> </p><p></p><p>=\frac{\sqrt3}{2}= </p><p>2</p><p>3</p><p> </p><p> </p><p> </p><p> </p><p></p><p>Therefore, \sec A\tan B-\cot A\sin B=\frac{\sqrt3}{2}secAtanB−cotAsinB= </p><p>2</p><p>3</p><p> </p><p>$

Step-by-step explanation:

• please make it a brainliest answer.