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Question: If \[\sin A = \dfrac{1}{{\sqrt {10} }}\]and \(\sin B = \dfrac{1}{{\sqrt 5 }}\), where A and B are po...

If sin⁡A=110\sin A = \dfrac{1}{{\sqrt {10} }}and sin⁡B=15\sin B = \dfrac{1}{{\sqrt 5 }}, where A and B are positive acute angles, then A + B is equal to
A.π\pi
B.π2\dfrac{\pi }{2}
C.π3\dfrac{\pi }{3}
D.π4\dfrac{\pi }{4}

Explanation

Solution

Hint : We are given the values of sin A and sin B. So, we will find the values of cos A and cos B using the appropriate trigonometric formula and the given values of sine. Then, we will put these values in a trigonometric formula which is cos(A+B) = cos A cos B – sin A sin B to find the value of A + B.

Complete step-by-step answer :
Given: sin⁡A=110\sin A = \dfrac{1}{{\sqrt {10} }}
sin⁡B=15\sin B = \dfrac{1}{{\sqrt 5 }}
A and B are acute angles.
Now we will find the value of cos A using the formula sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1.
Putting the value of sin A in the above formula.
(110)2+cos⁡2A=1{\left( {\dfrac{1}{{\sqrt {10} }}} \right)^2} + {\cos ^2}A = 1
cos⁡2A=1−110{\cos ^2}A = 1 - \dfrac{1}{{10}}
cos⁡2A=10−110{\cos ^2}A = \dfrac{{10 - 1}}{{10}}
cos⁡2A=910{\cos ^2}A = \dfrac{9}{{10}}
cos⁡A=310\cos A = \dfrac{3}{{\sqrt {10} }}.
Using the same formula we will find the value of cos B.
sin⁡2B+cos⁡2B=1{\sin ^2}B + {\cos ^2}B = 1
Putting the value of sin B in the above equation.
(15)2+cos⁡2B=1{\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2} + {\cos ^2}B = 1
cos⁡2B=1−15{\cos ^2}B = 1 - \dfrac{1}{5}
cos⁡2B=5−15{\cos ^2}B = \dfrac{{5 - 1}}{5}
cos⁡2B=45{\cos ^2}B = \dfrac{4}{5}
cos⁡B=25\cos B = \dfrac{2}{{\sqrt 5 }}
Now, we have to find values of A + B. So, we will let it equal to θ\theta .
A+B=θA + B = \theta
Taking cos on both sides.
cos⁡(A+B)=cos⁡θ\cos \left( {A + B} \right) = \cos \theta
cos⁡Acos⁡B−sin⁡Asin⁡B=θ\cos A\cos B - \sin A\sin B = \theta
Putting the values in the above equation.
310×25−110×15=cos⁡θ\dfrac{3}{{\sqrt {10} }} \times \dfrac{2}{{\sqrt 5 }} - \dfrac{1}{{\sqrt {10} }} \times \dfrac{1}{{\sqrt 5 }} = \cos \theta
6−110.5=cos⁡θ\dfrac{{6 - 1}}{{\sqrt {10} .\sqrt 5 }} = \cos \theta
cos⁡θ=550\cos \theta = \dfrac{5}{{\sqrt {50} }}
cos⁡θ=552\cos \theta = \dfrac{5}{{5\sqrt 2 }}
cos⁡θ=12\cos \theta = \dfrac{1}{{\sqrt 2 }}
Value of cos⁡θ\cos \theta is 12\dfrac{1}{{\sqrt 2 }}at π4\dfrac{\pi }{4}. So,
cos⁡θ=cos⁡π4\cos \theta = \cos \dfrac{\pi }{4}
θ=π4\theta = \dfrac{\pi }{4}
θ\theta is equal to A + B. So,
A+B=π4A + B = \dfrac{\pi }{4}.
So, option (4) is the correct answer.
So, the correct answer is “Option 4”.

Note : While finding the square root of the squared terms we have to consider both the positive and negative values. But in this case we will only consider the positive value of cos A and cos B because A and B are acute angles. So, they will lie in the first quadrant and the value of cos in the first quadrant is always positive.