Question

Prove the following:
$\cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \sin^{- 1} (\frac{56}{65})$

Solution

Let a be in I quadrant such that

$\cos^{- 1} (\frac{12}{13}) = a So \cos a = \frac{12}{13} \Rightarrow \sin a = \sqrt{1 - {(\frac{12}{13})}^{2}} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{169 - 144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} And \tan a = \frac{5}{12} So, a = \tan^{- 1} (\frac{5}{12}) . . . . . . . . . (i) Again b \in I quadrant such that \sin^{- 1} (\frac{3}{5}) = b So, \sin b = \frac{3}{5}$

$\Rightarrow \cos b = \sqrt{1 - {(\frac{3}{5})}^{2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} And \tan b = \frac{3}{4} so, b = \tan^{- 1} (\frac{3}{4}) . . . . . . . . . . . . . . . (ii) Now, let \sin^{- 1} (\frac{56}{65}) = c where c is in I quadrant So, \sin c = \frac{56}{65}$

$\Rightarrow \cos c = \sqrt{1 - {(\frac{56}{65})}^{2}} = \sqrt{1 - \frac{3136}{4225}} = \sqrt{\frac{4225 - 3136}{4225}} = \sqrt{\frac{1089}{4225}} = \frac{33}{65} And, \tan c = \frac{56}{33} So, c = \tan^{- 1} (\frac{56}{33}) \Rightarrow \sin^{- 1} (\frac{56}{65}) = \tan^{- 1} (\frac{56}{33}) . . . . . . . . . . . (iii) Now, we need to prove \cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \sin^{- 1} (\frac{56}{65})$

consider a + b

$= \cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{5}{12}) + \tan^{- 1} (\frac{3}{4}) [\begin{matrix} \cos^{- 1} (\frac{12}{13}) = \tan^{- 1} (\frac{5}{12}) and \\ \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{3}{4}) \end{matrix}] = \tan^{- 1} [\frac{\frac{5}{12} + \frac{3}{4}}{1 - (\frac{5}{12} x \frac{3}{4})}] [using, \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] = \tan^{- 1} (\frac{20 + 36}{48 - 15}) = \tan^{- 1} (\frac{56}{33}) = c = \sin^{- 1} (\frac{56}{65}) [using, eq, (iii)]$

Hence proved.

For Daily Free Study Material Join wiredfaculty

Inverse Trigonometric Functions

Prove the following:
$\cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \sin^{- 1} (\frac{56}{65})$

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location

For Daily Free Study Material Join wiredfaculty

Inverse Trigonometric Functions

Prove the following:cos-1 1213 + sin-1 35 = sin-1 5665

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Prove the following:
$\cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \sin^{- 1} (\frac{56}{65})$