Question

find the value of sin (a+b),cos (a+b)and tan (a+b),given: sina=3/5,cosb=5/13, a and b in quadrant 1​

Answer 1

Given that :-

sinA = 3/5

CosB = 5/13

A and B lie in first quadrant.

This means that all the trigonometric Functions will be positive as A and B are in First Quadrant.

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$\sin(a) = \frac{3}{5} \\ \\ { \sin}^{2}a = \frac{9}{25} \\ \\ \\ { \cosec}^{2}a = \frac{25}{9} \\ \\ \\1 + { \cot}^{2}a = \frac{25}{9} \\ \\ \\ { \cot}^{2}a = \frac{25}{9} - 1 = \frac{16}{9} \\ \\ \\ \cot a = \pm{ \frac{4}{3}}$

Since A is First Quadrant...

CotA = 4/3

Therefore :-

Tan A = 3/4

Now,

$\tan(a) = \frac{3}{4} \\ \\ \frac{ \sin(a) }{ \cos(a)} = \frac{3}{4} \\ \\ \\ \cos(a) = \frac{4 \sin(a) }{3} \\ \\ \\ \cos(a) = \frac{4 \times 3}{3 \times 5} = \frac{4}{5}$

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$\cos(b) = \frac{5}{13} \\ \\ { \cos}^{2}b = \frac{25}{169} \\ \\ \\ { \sec}^{2}b = \frac{169}{25} \\ \\ \\1 + { \tan}^{2}b = \frac{169}{25} \\ \\ \\ { \tan}^{2}b = \frac{169}{25} - 1 = \frac{144}{25} \\ \\ \\ \tan(b) = \frac{12}{5} \\ \\ \ \frac{\sin(b)}{\cos(b)} = \frac{12}{5} \\ \\ \sin(b) = \frac{12}{5} \times \frac{5}{13} = \frac{12}{13}$

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We have got the value of sinA, sinB, cosA, cosB, tanA, tanB

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sin(A+B) = sinAcosB+cosAsinB

Solving :-

$(\frac{3}{5} \times \frac{5}{13}) + (\frac{4}{5} \times \frac{12}{13}) \\ \\ = \frac{15 + 48}{65} = \frac{63}{65}$

cos(A+B) =cosAcosB-sinAsinB

$= \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13} \\ \\ = \frac{56}{65}$

tan(A+B) =

$\frac{\frac{3}{4} + \frac{12}{5}}{1 - \frac{9}{5}} \\ \\ \\ = \frac{ - 63}{16}$

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Answer 2

Answer:

Step-by-step explanation:

Given that :-

sinA = 3/5

CosB = 5/13

A and B lie in first quadrant.

This means that all the trigonometric Functions will be positive as A and B are in First Quadrant.