Question

Evaluate the expression: $\dfrac{{\sin \theta \cos \theta \sin ({{90}^0} - \theta )}}{{\cos ({{90}^0} - \theta )}} + \dfrac{{\cos \theta \sin \theta \cos ({{90}^0} - \theta )}}{{\sin ({{90}^0} - \theta )}} + \dfrac{{{{\sin }^2}{{27}^0} + {{\sin }^2}{{63}^0}}}{{{{\cos }^2}{{40}^0} + {{\cos }^2}{{50}^0}}}$
(a). 1
(b). 2
(c). 4
(d). 3

Answer 1

Hint- This question requires the basic concepts of Trigonometry and the Trigonometric Pythagorean identities. Use trigonometric ratios of angle $({90^0} - \theta )$ to proceed and simplify each expression to get the answer.

Complete step-by-step answer:

We have to evaluate $\dfrac{{\sin \theta \cos \theta \sin ({{90}^0} - \theta )}}{{\cos ({{90}^0} - \theta )}} + \dfrac{{\cos \theta \sin \theta \cos ({{90}^0} - \theta )}}{{\sin ({{90}^0} - \theta )}} + \dfrac{{{{\sin }^2}{{27}^0} + {{\sin }^2}{{63}^0}}}{{{{\cos }^2}{{40}^0} + {{\cos }^2}{{50}^0}}}$
We know that, $\sin ({90^0} - \theta ) = \cos \theta $ and $\cos ({90^0} - \theta ) = \sin \theta $
On substituting the values in the expression, we get
$ \Rightarrow \dfrac{{\sin \theta \cos \theta \cos \theta }}{{\sin \theta }} + \dfrac{{\cos \theta \sin \theta \sin \theta }}{{\cos \theta }} + \dfrac{{{{\sin }^2}{{27}^0} + {{\sin }^2}{{63}^0}}}{{{{\cos }^2}{{40}^0} + {{\cos }^2}{{50}^0}}}$
$ \Rightarrow {\cos ^2}\theta + {\sin ^2}\theta + \dfrac{{{{\sin }^2}{{27}^0} + {{\sin }^2}{{63}^0}}}{{{{\cos }^2}{{40}^0} + {{\cos }^2}{{50}^0}}}$
Again, we will use $\sin ({90^0} - \theta ) = \cos \theta $ and $\cos ({90^0} - \theta ) = \sin \theta $
Here, ${\sin ^2}{27^0} = {\sin ^2}({90^0} - {63^0}) = {\cos ^2}{63^0}$ and ${\cos ^2}{40^0} = {\cos ^2}({90^0} - {50^0}) = {\sin ^2}{50^0}$
Also, we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$ \Rightarrow 1 + \dfrac{{{{\cos }^2}{{63}^0} + {{\sin }^2}{{63}^0}}}{{{{\sin }^2}{{50}^0} + {{\cos }^2}{{50}^0}}}$
$ \Rightarrow 1 + \dfrac{1}{1}$
$ \Rightarrow 1 + 1 = 2$
Hence, our required expression evaluates down to 2
$\therefore $ Option B. 2 is our correct answer.

Note- Always remember in these types of questions, try to change the trigonometric ratios of angles $({90^0} \pm \theta )$ , $({180^0} \pm \theta )$ to the corresponding trigonometric ratios of angle $\theta $ and then apply the simple trigonometric identities like we used ${\sin ^2}\theta + {\cos ^2}\theta = 1$ here.