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Question: Find the value \({\sin ^{ - 1}}\dfrac{3}{5} + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right) = \) A....

Find the value sin⁡−135+tan⁡−1(17)={\sin ^{ - 1}}\dfrac{3}{5} + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right) =
A. π4\dfrac{\pi }{4}
B. π2\dfrac{\pi }{2}
C. cos⁡−1(45){\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right)
D. π\pi

Explanation

Solution

First, we need to analyze the given information so that we can able to solve the problem. Generally, in Mathematics, the trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Here in this question, we are asked to find the value ofsin⁡−135+tan⁡−1(17){\sin ^{ - 1}}\dfrac{3}{5} + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right).
We need to apply the appropriate trigonometric identities to obtain the required answer.
Formula to be used:
a) sin⁡θ=oppositehypotenuse\sin \theta = \dfrac{{opposite}}{{hypotenuse}}
b) tan⁡θ=oppositeadjacent\tan \theta = \dfrac{{opposite}}{{adjacent}}
c) The formula using Pythagorean Theorem is, c2=a2+b2{c^2} = {a^2} + {b^2} where cc is the hypotenuse, aa is the perpendicular (opposite), andbb is the base (adjacent)
d) tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)
e) tan⁡−1tan⁡x=x{\tan ^{ - 1}}\tan x = x

Complete step by step answer:
Letsin⁡−135=θ{\sin ^{ - 1}}\dfrac{3}{5} = \theta
⇒sin⁡θ=35\Rightarrow \sin \theta = \dfrac{3}{5}
Since we knowsin⁡θ=oppositehypotenuse\sin \theta = \dfrac{{opposite}}{{hypotenuse}}, we shall get the value of opposite and hypotenuse.
sin⁡θ=35\sin \theta = \dfrac{3}{5}
This implies, opposite=3 = 3 and hypotenuse=5 = 5
Using Pythagoras Theorem, we haveadjacent=hypotenuse2−opposite2adjacent = \sqrt {hypotenus{e^2} - opposit{e^2}}
Hence, adjacent=52−32= \sqrt {{5^2} - {3^2}}
=25−9 =4  = \sqrt {25 - 9} \\\ = 4 \\\
We need to calculatetan⁡θ\tan \theta using the formulatan⁡θ=oppositeadjacent\tan \theta = \dfrac{{opposite}}{{adjacent}}
tan⁡θ=34\tan \theta = \dfrac{3}{4}
⇒θ=tan⁡−1(34)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)
Hence, we getsin⁡−135+tan⁡−1(17){\sin ^{ - 1}}\dfrac{3}{5} + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right) =tan⁡−1(34)+tan⁡−1(17) = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right)
Now, we shall apply the formulatan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)in the above equation.
Hence, sin⁡−135+tan⁡−1(17){\sin ^{ - 1}}\dfrac{3}{5} + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right) =tan⁡−1(34+171−34×17) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{1}{7}}}{{1 - \dfrac{3}{4} \times \dfrac{1}{7}}}} \right)
=tan⁡−1(21+4281−328)= {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{21 + 4}}{{28}}}}{{1 - \dfrac{3}{{28}}}}} \right)
=tan⁡−1(21+42828−328)= {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{21 + 4}}{{28}}}}{{\dfrac{{28 - 3}}{{28}}}}} \right)
=tan⁡−1(25282528)= {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{25}}{{28}}}}{{\dfrac{{25}}{{28}}}}} \right)
=tan⁡−12528×2825= {\tan ^{ - 1}}\dfrac{{25}}{{28}} \times \dfrac{{28}}{{25}}
=tan⁡−11= {\tan ^{ - 1}}1
=tan⁡−1tan⁡π4= {\tan ^{ - 1}}\tan \dfrac{\pi }{4} (Here we applied1=tan⁡π41 = \tan \dfrac{\pi }{4})
=π4= \dfrac{\pi }{4} (Here we appliedtan⁡−1tan⁡x=x{\tan ^{ - 1}}\tan x = x )
Therefore, sin⁡−135+tan⁡−1(17)=π4{\sin ^{ - 1}}\dfrac{3}{5} + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right) = \dfrac{\pi }{4}

So, the correct answer is “Option A”.

Note: If we are asked to calculate the value of a trigonometric expression, we need to first analyze the given problem where we are able to apply the trigonometric identities.
Here, we have applied some trigonometric identities/formulae that are needed to know to obtain the desired answer. Hence, we gotsin⁡−135+tan⁡−1(17)=π4{\sin ^{ - 1}}\dfrac{3}{5} + {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right) = \dfrac{\pi }{4}.