If tan A -3-4- cos B -9-41- where - A -3 -2 and 0- B -2- find tan A - B -

We have, tanA= 3/4, cosB= 9/41 ∴ sinB= √(1 cos^2B) = √(1 ( 9/41 )^2) = √(1 81/1681) = √(1600/1681) = 40/41 ∴ tanB= sinB/cosB=40/41/9/41 = 40/9 Now, tan(A+B)=ta ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

If tan A =3/4...

Question

If $t a n A = \frac{3}{4}, c o s B = \frac{9}{41},$ where $π < A < \frac{3 π}{2}$ and $0 < B < \frac{π}{2},$ find $t a n (A + B)$ .

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Solution

We have,
$t a n A = \frac{3}{4}, c o s B = \frac{9}{41}$
$∴ s i n B = \sqrt{1 - c o s^{2} B}$
$= \sqrt{1 - {(\frac{9}{41})}^{2}}$
$= \sqrt{1 - \frac{81}{1681}}$
$= \sqrt{\frac{1600}{1681}}$
$= \frac{40}{41}$
$∴ t a n B = \frac{s i n B}{c o s B} = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{9}$
Now,
$t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$
$= \frac{\frac{3}{4} + \frac{40}{9}}{1 - \frac{3}{4} \times \frac{40}{9}} = \frac{\frac{27 + 160}{36}}{\frac{36 - 120}{36}}$ $= \frac{\frac{187}{36}}{\frac{- 84}{36}} = - \frac{187}{84}$

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If tanA=34,cosB=941, where π<A<3π2 and 0<B<π2, find tan(A+B).

We have, tanA=34,cosB=941 ∴sinB=√1−cos2B =√1−(941)2 =√1−811681 =√16001681 =4041 ∴tanB=sinBcosB=4041941=409 Now, tan(A+B)=tanA+tanB1−tanA tanB =34+4091−34×409=27+1603636−12036 =18736−8436=−18784

If $t a n A = \frac{3}{4}, c o s B = \frac{9}{41},$ where $π < A < \frac{3 π}{2}$ and $0 < B < \frac{π}{2},$ find $t a n (A + B)$ .