Question

if sin a =15/17, cos B=12/13, find the value of sin (a+b), cos (a-b) and tan (a+b) , where a, b are+ ve acute angles​

Answer 1

$\huge\boxed{\fcolorbox{black}{pink}{Answer}}$

Given

$\sin(A) = \frac{15}{17}$

$\cos(B) = \frac{12}{13}$

To Find

value of :-

$\sin(a + b)$

$\cos(a - b)$

$\tan(a + b)$

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Using sin²A + Cos²A =1 we get

◻ Cos²A = 1-sin²A

$◻ \: \cos ^{2} (A) = 1 - \frac{225}{289} = \frac{289 - 225}{289} = \frac{64}{289} = \frac{8}{17}$

• [Side can't negative ]

◻Sin²B = 1-cos²B

$◻ \sin ^{2} (B) = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169} = \frac{5}{13}$

• [Side can't negative ]

$\huge{therfore}$

$◻\sin(A + B) = \sin(A) \cos(B) + \cos(A) \sin(B)$

$◻ \frac{15}{17} \times \frac{12}{13} + \frac{8}{17} \times \frac{5}{13}$

$◻ \frac{180}{221} + \frac{40}{221}$

$◻ \frac{220}{221}$

$◻\cos(A - B) = \cos(A) \cos(B) - \sin(A) \sin(B)$

$◻ \frac{8}{17} \times \frac{12}{13} - \frac{15}{47} \times \frac{5}{13}$

$◻ \frac{96}{221} - \frac{75}{221}$

$◻ \frac{21}{221}$

$◻ \tan(A) = \frac{ \sin(A) }{A}$

$◻ \frac{ \frac{15}{17} }{ \frac{8}{17} } = \frac{15}{8}$

$◻ \tan(B) = \frac{ \sin(B) }{ \cos(B) }$

$◻ \frac{ \frac{5}{13} }{ \frac{12}{13} } = \frac{5}{12}$

$◻ \tan(A + B)$

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

$◻ \frac{( \frac{15}{8} + \frac{15}{12}) }{1 - ( \frac{15}{8} )( \frac{15}{12}) }$

$◻ \frac{( \frac{ \frac{45}{10}}{24} )}{( \frac{ \frac{96}{75} }{96}) } = \frac{55}{24} \times \frac{96}{21} = \frac{210}{21} = 10$