If sin A -sin B and cos A -cos B- then-EAMCET 1994-A- sinA-B-2-0B- sinA-B-2-0C- cosA-B-2-0D- cos A-B-0

The correct option is A We have sin A = sin B and cos A = cos B 2cos ( A+B/2 )sin ( A B/2 )=0 nbsp;, 2sin ( A+B/2 )sin ( A B/2 )=0 Hence, sin( A B/2) = 0 ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

If sin A =sin...

Question

If sinA=sinB and cosA=cosB, then

[EAMCET 1994]

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Solution

The correct option is A

We have sin A = sin B and cos A = cos B

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

Hence, sin $(\frac{A - B}{2})$ = 0.

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Similar questions

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}

Q. If

sin A = sin B

and

cos A = cos B

, then

sin \frac{A - B}{2} =

Prove that:
(i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$
(ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$
(iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$
(iv) $\frac{s i n A + s i n B}{s i n A - s i n B} t a n (\frac{A + B}{2}) c o t \frac{A + B}{2}$
(v) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$

If cos(A-B) = 3/5 and tanA*tanB = 2 then

(a) cosA*cosB= 1/5 (c) cos(A+B)= -1/5

(b) sinA*sinB= 2/5 (d) sinA*sinB=4/5

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}

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If sinA=sinB and cosA=cosB, then [EAMCET 1994]

The correct option is A We have sin A = sin B and cos A = cos B 2cos(A+B2)sin(A−B2)=0,−2sin(A+B2)sin(A−B2)=0 Hence, sin(A−B2) = 0.

If sinA=sinB and cosA=cosB, then

[EAMCET 1994]

The correct option is A

We have sin A = sin B and cos A = cos B

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

Hence, sin $(\frac{A - B}{2})$ = 0.