If cos A - B -3-5 and tan A and tan B -2- thenA- cos A cos B-1-5B- cos A cos B-1-5C- sin A sin B -1-5D- sin A sin B-1-5

The correct option is A cosA cosB= 1/5 cosA cosB= 1/5 tanA tanB= sinA sinB/cosA cosB =2 (given) ...(1) Also, cos(A B)= 3/5 ⇒ cosAcosB+sinAsinB= 3/5 ∴ sinAsin ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If cos A - B ...

Question

If $c o s (A - B) = \frac{3}{5}$ and tanA and tanB=2, then

c o s A c o s B = \frac{1}{5}

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c o s A c o s B = - \frac{1}{5}

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s i n A s i n B = - \frac{1}{5}

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s i n A s i n B = - \frac{1}{5}

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Solution

The correct option is A $c o s A c o s B = \frac{1}{5}$

cosA cosB= \frac{1}{5}
$t a n A t a n B = \frac{s i n A s i n B}{c o s A c o s B} = 2$ (given) ...(1)
Also,
$c o s (A - B) = \frac{3}{5}$
$\Rightarrow c o s A c o s B + s i n A s i n B = \frac{3}{5}$
$∴ s i n A s i n B = \frac{3}{5} - c o s A c o s B$ ...(2)
Substituting eq.(2) in eq.(1), we get:
$\Rightarrow \frac{\frac{3}{5} - c o s A c o s B}{c o s A c o s B} = 2$
$\Rightarrow 3 c o s A c o s B = \frac{3}{5}$
$\Rightarrow c o s A c o s B = \frac{1}{5}$

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Similar questions

If cos (A-B) = $\frac{3}{5}$ and tanAtanB = 2,then

Q. If

tan A = 1

and

tan B = \sqrt{3}

evaluate:

cos A cos B - sin A sin B

Q. If cos (A − B)

= \frac{3}{5}

and tan A tan B = 2, then

(a)

\cos A \cos B = \frac{1}{5}

(b)

\cos A \cos B = - \frac{1}{5}

(c)

\sin A \sin B = - \frac{1}{5}

(d)

\sin A \sin B = - \frac{1}{5}

Q. If

A = B = 60^{\circ}

, verify that
(i)

cos (A - B) = cos A cos B + sin A sin B

(ii)

sin (A - B) = sin A cos B - cos A sin B

(iii)

tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

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

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If cos(A−B)=35 and tanA and tanB=2, then

The correct option is A cosAcosB=15 cosA cosB= \frac{1}{5} tanAtanB=sinAsinBcosAcosB=2 (given) ...(1) Also, cos(A−B)=35 ⇒cosAcosB+sinAsinB=35 ∴sinAsinB=35−cosAcosB ...(2) Substituting eq.(2) in eq.(1), we get: ⇒35−cosAcosBcosAcosB=2 ⇒3cosAcosB=35 ⇒cosAcosB=15

If $c o s (A - B) = \frac{3}{5}$ and tanA and tanB=2, then