Question

What is the value of expression $\dfrac{{(2\tan 4 + 3\cot 4)(2\cot 4 + 3\tan 4)}}{{24{{\cot }^2}8 + 25}}$ is,
A. 1
B. 2
C. 3
D. 4

Answer 1

Start by converting all the trigonometric quantities in terms of sin and cos ,Take L.C.M. and simplify . Use the trigonometric identities and formulas for ease of calculation and simplification .Re -arrange the terms , so that it becomes easy to recognize and solve.

Complete step-by-step answer:
In this question, it is given to us that we have to find the value of the expression, $\dfrac{{(2\tan 4 + 3\cot 4)(2\cot 4 + 3\tan 4)}}{{24{{\cot }^2}8 + 25}}$
Expressing the above in terms of sin, cos ,we get
$\dfrac{{(2\dfrac{{\sin 4}}{{\cos 4}} + 3\dfrac{{\cos 4}}{{\sin 4}})(2\dfrac{{\cos 4}}{{\sin 4}} + 3\dfrac{{\sin 4}}{{\cos 4}})}}{{24{{\left( {\dfrac{{\cos 8}}{{\sin 8}}} \right)}^2} + 25}}$
Taking L.C.M and simplifying , we get
$\dfrac{{(\dfrac{{2{{\sin }^2}4 + 3{{\cos }^2}4}}{{\cos 4\sin 4}})(\dfrac{{2{{\cos }^2}4 + 3{{\sin }^2}4}}{{\sin 4\cos 4}})}}{{\dfrac{{24{{\cos }^2}8 + 25{{\sin }^2}8}}{{{{\sin }^2}8}}}}$
Now , we can break few of the terms in order to simplify more, we get
$\dfrac{{\left( {\dfrac{{2{{\sin }^2}4 + 2{{\cos }^2}4 + {{\cos }^2}4}}{{\cos 4\sin 4}}} \right)\left( {\dfrac{{2{{\cos }^2}4 + 2{{\sin }^2}4 + {{\sin }^2}4}}{{\sin 4\cos 4}}} \right)}}{{\dfrac{{24{{\cos }^2}8 + 24{{\sin }^2}8 + {{\sin }^2}8}}{{{{\sin }^2}8}}}}$
We know, ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\dfrac{{\left( {\dfrac{{2 + {{\cos }^2}4}}{{\cos 4\sin 4}}} \right)\left( {\dfrac{{2 + {{\sin }^2}4}}{{\sin 4\cos 4}}} \right)}}{{\dfrac{{24 + {{\sin }^2}8}}{{{{\sin }^2}8}}}}$
Now , further breaking sin 8 as sin 2(4) , we get
$\dfrac{{\left( {\dfrac{{2 + {{\cos }^2}4}}{{\cos 4\sin 4}}} \right)\left( {\dfrac{{2 + {{\sin }^2}4}}{{\sin 4\cos 4}}} \right)}}{{\dfrac{{24 + {{\sin }^2}2(4)}}{{{{\sin }^2}2(4)}}}}$
Re- arranging the numerator and denominators , we get
$\dfrac{{\left( {2 + {{\cos }^2}4} \right)\left( {2 + {{\sin }^2}4} \right)\left( {{{\sin }^2}2(4)} \right)}}{{\left( {24 + {{\sin }^2}2(4)} \right)\left( {{{\cos }^2}4{{\sin }^2}4} \right)}}$
Now , we will use the formula $\sin 2\theta = 2\sin \theta \cos \theta$
$\dfrac{{\left( {2 + {{\cos }^2}4} \right)\left( {2 + {{\sin }^2}4} \right){{\left( {2\sin 4\cos 4} \right)}^2}}}{{\left( {24 + 4{{\sin }^2}4{{\cos }^2}4} \right)\left( {{{\cos }^2}4{{\sin }^2}4} \right)}}$
By simplification, we will get:
$\dfrac{{\left( {2 + {{\cos }^2}4} \right)\left( {2 + {{\sin }^2}4} \right)4}}{{\left( {24 + 4{{\sin }^2}4{{\cos }^2}4} \right)}}$
By multiplying, we will get
$\dfrac{{16 + 8{{\sin }^2}4 + 8{{\cos }^2}4 + 4{{\sin }^2}4{{\cos }^2}4}}{{\left( {24 + 4{{\sin }^2}4{{\cos }^2}4} \right)}} \\\ \Rightarrow \dfrac{{16 + 8({{\sin }^2}4 + {{\cos }^2}4) + 4{{\sin }^2}4{{\cos }^2}4}}{{\left( {24 + 4{{\sin }^2}4{{\cos }^2}4} \right)}} \\\ \\\$
Using the identity, ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\Rightarrow \dfrac{{16 + 8 + 4{{\sin }^2}4{{\cos }^2}4}}{{\left( {24 + 4{{\sin }^2}4{{\cos }^2}4} \right)}} \\\ \Rightarrow \dfrac{{24 + 4{{\sin }^2}4{{\cos }^2}4}}{{\left( {24 + 4{{\sin }^2}4{{\cos }^2}4} \right)}} \\\ = 1 \\\ \\\$

So, the correct answer is “Option A”.

Note: In this particular question, the most used formula is $\sin 2\theta = 2\sin \theta \cos \theta$. It should be noted that in such a type of question we have to use trigonometric identities like ${\sin ^2}\theta + {\cos ^2}\theta = 1$ . By using these basics one can easily solve these types of questions.