Question

Evaluate:$\dfrac{\sec {{29}^{\circ }}}{\cos ec{{61}^{\circ }}}+2\cot {{8}^{\circ }}\cot {{17}^{\circ }}\cot {{45}^{\circ }}\cot {{73}^{\circ }}\cot {{82}^{\circ }}-3\left( {{\sin }^{2}}{{38}^{\circ }}+{{\sin }^{2}}{{52}^{\circ }} \right)$.

Answer 1

Hint: We solve the whole equation using the
sin (90° - θ) = cos θ
cos (90° - θ) = sin θ.
tan (90° - θ) = cot θ.
cosec (90° - θ) = sec θ.
sec (90° - θ) = cosec θ.
cot (90° - θ) = tan θ.
The above all properties help us in solving the question.

Complete step-by-step answer:
$\dfrac{\sec {{29}^{\circ }}}{\cos ec{{61}^{\circ }}}+2\cot {{8}^{\circ }}\cot {{17}^{\circ }}\cot {{45}^{\circ }}\cot {{73}^{\circ }}\cot {{82}^{\circ }}-3({{\sin }^{2}}{{38}^{\circ }}+{{\sin }^{2}}{{52}^{\circ }})$
So, now we eliminate the variables whose value is easily evaluated. In this part $\cot 45{}^\circ =1$, so we put 1 in the place of cot and then our expression reduces to:
$\dfrac{\sec {{29}^{\circ }}}{\operatorname{co}\sec {{61}^{\circ }}}+2\cot {{8}^{\circ }}\cot {{17}^{\circ }}\left( 1 \right)\cot 73{}^\circ \cot 82{}^\circ -3({{\sin }^{2}}{{38}^{\circ }}+{{\sin }^{2}}{{52}^{\circ }})$
Now rewriting cosec term: cosec (90° - θ) = sec θ
$\dfrac{\sec {{29}^{\circ }}}{\sec {{29}^{\circ }}}+2\cot {{8}^{\circ }}\cot {{17}^{\circ }}\left( 1 \right)\cot 73{}^\circ \cot 82{}^\circ -3({{\sin }^{2}}{{38}^{\circ }}+{{\sin }^{2}}{{52}^{\circ }})$
Again, rewriting the cot term: cot (90° - θ) = tan θ.
$\dfrac{\sec {{29}^{\circ }}}{\sec {{29}^{\circ }}}+2\cot {{8}^{\circ }}\cot {{17}^{\circ }}\left( 1 \right)\cot (90-17){}^\circ \cot (90-8){}^\circ -3({{\sin }^{2}}{{38}^{\circ }}+{{\sin }^{2}}{{52}^{\circ }})$
Converting cot θ to tan θ we get,
$\dfrac{\sec {{29}^{\circ }}}{\sec {{29}^{\circ }}}+2\cot {{8}^{\circ }}\cot {{17}^{\circ }}\left( 1 \right)\tan 17{}^\circ \tan 8{}^\circ -3({{\sin }^{2}}{{38}^{\circ }}+{{\sin }^{2}}{{52}^{\circ }})$
Now, cancelling the cot and tan terms as they are reciprocal trigonometric identities we get,
$\dfrac{\sec {{29}^{\circ }}}{\sec {{29}^{\circ }}}+2-3({{\sin }^{2}}{{38}^{\circ }}+{{\sin }^{2}}{{52}^{\circ }})$
Cancelling sec θ in the first part we get,
$1+2-3({{\sin }^{2}}{{38}^{\circ }}+{{\sin }^{2}}{{52}^{\circ }})$
Converting sine terms: sin (90° - θ) = cos θ
$\begin{aligned} & 3-3({{\sin }^{2}}38{}^\circ +{{\sin }^{2}}(90-52){}^\circ \\\ & 3-3({{\sin }^{2}}38{}^\circ +{{\cos }^{2}}38{}^\circ ) \\\ \end{aligned}$
Now, using the identity of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we can reduce the last term as,
$\begin{aligned} & 3-3(1) \\\ & 3-3=0 \\\ \end{aligned}$
So, the final value obtained after performing a series of operations is 0.
Therefore, the answer obtained is 0.

Note: The key step is to rewrite all the variables so that each gets cancelled out using its counterpart and hence we get a final simple numerical evaluation.
All the trigonometric identities must be remembered to solve the question.