Question

Evaluate: $\dfrac{{3\cos {{53}^o}{\text{cosec}}{{37}^o}}}{{\left( {{{\cos }^2}{{29}^o} + {{\cos }^2}{{61}^o}} \right)}} - 3{\tan ^2}{45^o}$
$\left( a \right)$ 1
$\left( b \right)$ 3
$\left( c \right)$ 6
$\left( d \right)$ 0

Answer 1

Hint: In this particular question use the concept that cos (90 – x) = sin x so by this property first simplify the given trigonometric equation then again use the basic trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$ so use these concepts to reach the solution of the question.

Complete step by step answer:
Given trigonometric equation is $\dfrac{{3\cos {{53}^o}{\text{cosec}}{{37}^o}}}{{\left( {{{\cos }^2}{{29}^o} + {{\cos }^2}{{61}^o}} \right)}} - 3{\tan ^2}{45^o}$
The above equation is also written as
$ \Rightarrow \dfrac{{3\cos \left( {{{90}^o} - {{37}^o}} \right){\text{cosec}}{{37}^o}}}{{\left( {{{\cos }^2}\left( {{{90}^o} - {{61}^o}} \right) + {{\cos }^2}{{61}^o}} \right)}} - 3{\tan ^2}{45^o}$
Now as we know that cos (90 – x) = sin x, so use this property in the above equation we have,
$ \Rightarrow \dfrac{{3\sin {{37}^o}{\text{cosec}}{{37}^o}}}{{\left( {{{\sin }^2}{{61}^o} + {{\cos }^2}{{61}^o}} \right)}} - 3{\tan ^2}{45^o}$
Now as we know that ${\sin ^2}x + {\cos ^2}x = 1$ so use this property in the above equation we have,
$ \Rightarrow \left( {{{\sin }^2}{{61}^o} + {{\cos }^2}{{61}^o}} \right) = 1$
$ \Rightarrow \dfrac{{3\sin {{37}^o}{\text{cosec}}{{37}^o}}}{{\left( 1 \right)}} - 3{\tan ^2}{45^o}$
$ \Rightarrow 3\sin {37^o}{\text{cosec}}{37^o} - 3{\tan ^2}{45^o}$
Now as we know that sin x = (1/cosec x) so use this property in the above equation we have,
$ \Rightarrow 3\dfrac{{{\text{cosec}}{{37}^o}}}{{{\text{cosec}}{{37}^o}}} - 3{\tan ^2}{45^o}$
Now cancel out the common terms from numerator and denominator we have,
$ \Rightarrow 3 - 3{\tan ^2}{45^o}$
Now we also know that the value of tan 45 = 1, so use this value in the above equation we have,
$ \Rightarrow 3 - 3{\left( 1 \right)^2} = 3 - 3 = 0$
So the value of the given trigonometric equation is 0.
$ \Rightarrow \dfrac{{3\cos {{53}^o}{\text{cosec}}{{37}^o}}}{{\left( {{{\cos }^2}{{29}^o} + {{\cos }^2}{{61}^o}} \right)}} - 3{\tan ^2}{45^o} = 0$
So this is the required answer.
Hence option (D) is the correct answer.


Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric properties which are stated above then first simplify the given equation using these properties as above, then using the value of standard tan angle i.e. tan 45 = 1, and simplify we will get the required answer.