If sin-13-5 - cos-1 12-13 - sin-1 C- then C -

Given sin^ 1C = sin^ 13/5 + cos^ 112/13 ⇒ C = sin ( sin^ 13/5 + cos^ 112/13 ) Using sin^(A + B) = sin A cos B+ cos A sin B ⇒ C = 3/5×12/13 + √(1 9/25)√(1 ...

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Standard IX

History

Machiavelli

If sin-13/5 +...

Question

If $s i n^{- 1} \frac{3}{5} + c o s^{- 1} (\frac{12}{13}) = s i n^{- 1} C, t h e n C =$

\frac{65}{56}

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\frac{24}{65}

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\frac{16}{65}

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\frac{56}{65}

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Solution

The correct option is D $\frac{56}{65}$
$G i v e n s i n^{- 1} C = s i n^{- 1} \frac{3}{5} + c o s^{- 1} \frac{12}{13} \Rightarrow C = s i n (s i n^{- 1} \frac{3}{5} + c o s^{- 1} \frac{12}{13}) U s i n g s i n^{(} A + B) = s i n A c o s B + c o s A s i n B \Rightarrow C = \frac{3}{5} \times \frac{12}{13} + \sqrt{1 - \frac{9}{25}} \sqrt{1 - \frac{144}{169}} \Rightarrow C = \frac{56}{65} .$

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Similar questions

Q. Prove:

{cos}^{- 1} \frac{12}{13} + {sin}^{- 1} \frac{3}{5} = {sin}^{- 1} \frac{56}{65}

Q. Prove that

{s i n}^{- 1} (\frac{3}{5}) + {c o s}^{- 1} (\frac{12}{13}) = {s i n}^{- 1} (\frac{56}{65})

Q. Prove that :

{s i n}^{- 1} \frac{3}{5} - {c o s}^{- 1} \frac{12}{13} = {s i n}^{- 1} \frac{16}{65}

{sin}^{- 1} \frac{3}{5} + {cos}^{- 1} \frac{12}{13} = {cos}^{- 1} \frac{33}{65}

Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

Prove

(a)

{s i n}^{- 1} \frac{4}{5} + {s i n}^{- 1} \frac{5}{13} + {s i n}^{- 1} \frac{16}{65} = \frac{π}{2}

(b)

{s i n}^{- 1} \frac{3}{5} + {s i n}^{- 1} \frac{8}{17} = {c o s}^{- 1} \frac{36}{85}

(c)

{s i n}^{- 1} \frac{3}{5} + {c o s}^{- 1} \frac{12}{13} = {c o s}^{- 1} \frac{33}{65}

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If sin−135+cos−1(1213)=sin−1 C,then C=

The correct option is D 5665Given sin−1C=sin−135+cos−11213⇒C=sin(sin−135+cos−11213)Using sin(A+B)=sin A cos B+cos A sin B⇒C=35×1213+√1−925√1−144169⇒C=5665.

If $s i n^{- 1} \frac{3}{5} + c o s^{- 1} (\frac{12}{13}) = s i n^{- 1} C, t h e n C =$