If A- B- C are in A-P- then sin A-sin C-cos C-cosA-

The correct option is D cot B If A, B, C are in A.P. B A=C B Or. 2B=A+C=sinA sinC/cosC cosA =2sin ( A C/2 )cos ( A+C/2 )/ 2sin ( C+A/2 )sin ( C A/2 ) [ ∵ sin ...

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

Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If A, B, C ar...

Question

If A, B, C are in A.P., then $\frac{s i n A - s i n C}{c o s C - c o s A} =$

tanB

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

cot B

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

tan 2B

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

None of these

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

Open in App

Solution

The correct option is B
cot B

If A, B, C are in A.P.
B-A=C-B
Or. $2 B = A + C = \frac{s i n A - s i n C}{c o s C - c o s A} = \frac{2 s i n (\frac{A - C}{2}) c o s (\frac{A + C}{2})}{- 2 s i n (\frac{C + A}{2}) s i n (\frac{C - A}{2})} [∵ s i n A - s i n B = 2 s i n (\frac{A - B}{2}) c o s (\frac{A + B}{2})] a n d c o s A - c o s B = - 2 s i n (\frac{A + B}{2}) c o s (\frac{A - B}{2}) = \frac{s i n (\frac{A - C}{2}) c o s (\frac{A + C}{2})}{s i n (\frac{A + C}{2}) s i n (\frac{A - C}{2})} = \frac{c o s (\frac{A + C}{2})}{s i n (\frac{A + C}{2})} = \frac{c o s B}{s i n B} = c o t B$

Suggest Corrections

Similar questions

Q. If three angles A,B,C are in A.P then the value of

\frac{sin A - sin C}{cos C - cos A}

is equal to -

Q. If

sin A + sin B + sin C = 0

and

cos A + cos B + cos C = 0

then

cos (A + B) + cos (B + C) + cos (C + A) =

Q. If

cos a + cos b + cos c = sin a + sin b + sin c = 0

then

cos (a - b) + cos (b - c) + cos (c - a) = - \frac{3}{2}

Q. In any

Δ A B C, \frac{a + b}{c} =

Q. In an acute angles triangle

A B C

cos A, 1 - cos B, cos C

are in

A . P .

and

sin A + sin C = 1

, then

∠ B

is equal to

Join BYJU'S Learning Program

If A, B, C are in A.P., then sinA−sinCcosC−cosA=

The correct option is B cot B If A, B, C are in A.P. B-A=C-B Or. 2B=A+C=sinA−sinCcosC−cosA=2sin(A−C2)cos(A+C2)−2sin(C+A2)sin(C−A2)[∵sinA−sinB=2sin(A−B2)cos(A+B2)]and cosA−cosB=−2sin(A+B2)cos(A−B2)=sin(A−C2)cos(A+C2)sin(A+C2)sin(A−C2)=cos(A+C2)sin(A+C2)=cosBsinB=cotB

If A, B, C are in A.P., then $\frac{s i n A - s i n C}{c o s C - c o s A} =$