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Question: Show that \[{\sin ^{ - 1}}\dfrac{{12}}{{13}} + {\cos ^{ - 1}}\dfrac{4}{5} + {\tan ^{ - 1}}\dfrac{{63...

Show that sin⁡−11213+cos⁡−145+tan⁡−16316=π{\sin ^{ - 1}}\dfrac{{12}}{{13}} + {\cos ^{ - 1}}\dfrac{4}{5} + {\tan ^{ - 1}}\dfrac{{63}}{{16}} = \pi

Explanation

Solution

Here we will convert all the quantities in terms of tan⁡−1θ{\tan ^{ - 1}}\theta first using the trigonometric ratios and then use the following identity to get the desired solution.
tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)
We will also make use of the pythagoras theorem in the solution.

Complete step-by-step answer:
Let us first consider the left hand side:
LHS=sin⁡−11213+cos⁡−145+tan⁡−16316LHS = {\sin ^{ - 1}}\dfrac{{12}}{{13}} + {\cos ^{ - 1}}\dfrac{4}{5} + {\tan ^{ - 1}}\dfrac{{63}}{{16}}………………………. (1)
We know that,
θ=sin⁡−1(perpendicularhypotenuse)\theta = {\sin ^{ - 1}}\left( {\dfrac{{perpendicular}}{{hypotenuse}}} \right)
Hence, perpendicular=12perpendicular = 12
hypotenuse=13hypotenuse = 13
Now according to Pythagoras theorem we know that,
(hypotenuse)2=(base)2+(perpendicular)2{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {perpendicular} \right)^2}
Putting in the respective values we get:-
(13)2=(base)2+(12)2{\left( {13} \right)^2} = {\left( {base} \right)^2} + {\left( {12} \right)^2}
Simplifying it further we get:-
(base)2=169−144{\left( {base} \right)^2} = 169 - 144
⇒(base)2=25\Rightarrow {\left( {base} \right)^2} = 25
Taking square root we get:-
base=5base = 5
Now we know that,
θ=tan⁡−1(perpendicularbase)\theta = {\tan ^{ - 1}}\left( {\dfrac{{perpendicular}}{{base}}} \right)
Putting the values we get:-
θ=tan⁡−1(125)\theta = {\tan ^{ - 1}}\left( {\dfrac{{12}}{5}} \right)…………………….. (2)
Now we know that,
α=cos⁡−1(basehypotenuse)\alpha = {\cos ^{ - 1}}\left( {\dfrac{{base}}{{hypotenuse}}} \right)
Therefore,
base=4base = 4
hypotenuse=5hypotenuse = 5
Now according to Pythagoras theorem we know that,
(hypotenuse)2=(base)2+(perpendicular)2{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {perpendicular} \right)^2}
Putting in the respective values we get:-
(5)2=(4)2+(perpendicular)2{\left( 5 \right)^2} = {\left( 4 \right)^2} + {\left( {perpendicular} \right)^2}
Simplifying it further we get:-
(perpendicular)2=25−16{\left( {perpendicular} \right)^2} = 25 - 16
⇒(perpendicular)2=9\Rightarrow {\left( {perpendicular} \right)^2} = 9
Taking square root we get:-
perpendicular=3perpendicular = 3
Now we know that,
α=tan⁡−1(perpendicularbase)\alpha = {\tan ^{ - 1}}\left( {\dfrac{{perpendicular}}{{base}}} \right)
Putting the values we get:-
α=tan⁡−1(34)\alpha = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)……………………………. (3)
Now putting the values from equation 2 and equation 3 in equation 1 we get:-
LHS=tan⁡−1(125)+tan⁡−1(34)+tan⁡−16316LHS = {\tan ^{ - 1}}\left( {\dfrac{{12}}{5}} \right) + {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}
Now applying the following identity for first two terms :-
tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)
We get:-
LHS=tan⁡−1(125+341−(125)(34))+tan⁡−16316LHS = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{12}}{5} + \dfrac{3}{4}}}{{1 - \left( {\dfrac{{12}}{5}} \right)\left( {\dfrac{3}{4}} \right)}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}
Taking the LCM we get:-
LHS=tan⁡−1(12(4)+3(5)201−3620)+tan⁡−16316LHS = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{12\left( 4 \right) + 3\left( 5 \right)}}{{20}}}}{{1 - \dfrac{{36}}{{20}}}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}
Again taking the LCM in denominator we get:-
LHS=tan⁡−1(48+152020−3620)+tan⁡−16316LHS = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{48 + 15}}{{20}}}}{{\dfrac{{20 - 36}}{{20}}}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}
Simplifying it further we get:-
LHS=tan⁡−1(63−16)+tan⁡−16316LHS = {\tan ^{ - 1}}\left( {\dfrac{{63}}{{ - 16}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}
This implies,
LHS=tan⁡−1(−6316)+tan⁡−16316LHS = {\tan ^{ - 1}}\left( { - \dfrac{{63}}{{16}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}
Again applying the following identity:-
tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)
We get:-
LHS=tan⁡−1(−6316+63161−(−6316)(6316))LHS = {\tan ^{ - 1}}\left( {\dfrac{{ - \dfrac{{63}}{{16}} + \dfrac{{63}}{{16}}}}{{1 - \left( { - \dfrac{{63}}{{16}}} \right)\left( {\dfrac{{63}}{{16}}} \right)}}} \right)
Simplifying it we get:-
LHS=tan⁡−1(01−(−1))LHS = {\tan ^{ - 1}}\left( {\dfrac{0}{{1 - \left( { - 1} \right)}}} \right)
⇒LHS=tan⁡−1(02)\Rightarrow LHS = {\tan ^{ - 1}}\left( {\dfrac{0}{2}} \right)
⇒LHS=tan⁡−1(0)\Rightarrow LHS = {\tan ^{ - 1}}\left( 0 \right)
Now we know that,
tan⁡π=0\tan \pi = 0
⇒tan⁡−10=π\Rightarrow {\tan ^{ - 1}}0 = \pi
Hence, substituting the value we get:-
⇒LHS=π\Rightarrow LHS = \pi
Now considering RHS we get:-
RHS=πRHS = \pi
Hence, LHS=RHSLHS = RHS
Hence, proved.

Note: Students should use the trigonometric ratios carefully and convert them using the Pythagoras theorem.
In a right angled triangle,
(hypotenuse)2=(base)2+(perpendicular)2{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {perpendicular} \right)^2}