If tan A -1-7 and tan B -1-3- show that cos 2 A -sin 4 B-

We have, tan A=1/7 and tan B=1/3 ∴ cos 2A =1 tan^2 A/1+tan^2 A = 1 (1/7)^2/1+ (1/7 )^2=48/49/50/49 = 48/50=24/25 . . .(A) Also, sin 4B = sin2.2B = ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If tan A =1/7...

Question

If $t a n A = \frac{1}{7} a n d t a n B = \frac{1}{3}$ , show that cos 2A = sin 4B.

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Solution

We have,

$t a n A = \frac{1}{7} a n d t a n B = \frac{1}{3}$

$∴ c o s 2 A = \frac{1 - t a n^{2} A}{1 + t a n^{2} A} = \frac{1 - {(\frac{1}{7})}^{2}}{1 + {(\frac{1}{7})}^{2}} = \frac{48}{\frac{49}{\frac{50}{49}}}$
$= \frac{48}{50} = \frac{24}{25}$ . . .(A)

Also,

sin 4B = sin2.2B

= 2 sin 2B. cos 2B

$= 2. (\frac{2 t a n B}{1 + t a n^{2} B}) . (\frac{1 - t a n^{2} B}{1 + t a n^{2} B}) = 4. (\frac{\frac{1}{3}}{1 + \frac{1}{9}}) . (\frac{1 - \frac{1}{9}}{1 + \frac{1}{9}}) = \frac{4. \frac{1}{3} . \frac{8}{9}}{\frac{10}{9} \times \frac{10}{9}} = \frac{32 \times 3}{100}$
$= \frac{8 \times 3}{25} = \frac{24}{25}$ . . .(B)

from (A) and (B)

cos2A = sin 4B

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If tan A=17 and tan B=13, show that cos 2A = sin 4B.

We have, tan A=17 and tan B=13 ∴ cos 2A=1−tan2 A1+tan2 A=1−(17)21+(17)2=48495049 =4850=2425 . . .(A) Also, sin 4B = sin2.2B = 2 sin 2B. cos 2B =2.(2 tan B1+tan2B).(1−tan2B1+tan2B)=4.(131+19).(1−191+19)=4.13.89109×109=32×3100 =8×325=2425 . . .(B) from (A) and (B) cos2A = sin 4B

If $t a n A = \frac{1}{7} a n d t a n B = \frac{1}{3}$ , show that cos 2A = sin 4B.