Sin A-sin B-cos A-cosB-cos A-cos B-sin A-sin B-

The correct option is B 0 sinA sinB/cosA + cosB + cosA cosB/sinA + sinB = (sinA sinB) (sinA + sinB) + (cosA cosB) (cosA + cosB)/(cosA + cosB) (sinA + ...

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

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

sin A-sin B/c...

Question

$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B} =$

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
0

$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B}$

= $\frac{(s i n A - s i n B) (s i n A + s i n B) + (c o s A - c o s B) (c o s A + c o s B)}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{s i n^{2} A - s i n^{2} B + c o s^{2} A - c o s^{2} B}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{s i n^{2} A + c o s^{2} A - (s i n^{2} B ‘ + c o s^{2} B)}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{1 - 1}{(c o s A + c o s B) (s i n A + s i n B)}$
= 0

Suggest Corrections

Similar questions

Q. Prove that :

\frac{sin A - sin B}{cos A + cos B} + \frac{cos A - cos B}{sin A + sin B} = 0

Simplify the expression:
$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B}$

{(\frac{c o s A + c o s B}{s i n A - s i n B})}^{2014} + {(\frac{s i n A + s i n B}{c o s A - c o s B})}^{2014} = . . . . . . . . . . .

sin a + sin b = - \frac{21}{65}, cos a + cos b = - \frac{27}{65} \Rightarrow cos (\frac{a - b}{2}) =

Q. Assertion :

{(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n}

= 2 {cot}^{n} (\frac{A - B}{2})

n

is even

= 0

n

is odd. Reason:

\frac{cos A + cos B}{sin A - sin B} = cot \frac{A - B}{2}

Join BYJU'S Learning Program

sinA−sinBcosA+cosB+cosA−cosBsinA+sinB=

$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B} =$