Question

Solve and find the value of: $\dfrac{{\dfrac{{2\sin {{140}^ \circ }\sec {{280}^ \circ }}}{{\sec {{220}^ \circ }}} + \dfrac{{\sec {{340}^ \circ }}}{{\cos ec{{20}^ \circ }}}}}{{\dfrac{{\cot {{200}^ \circ } - \tan {{280}^ \circ }}}{{\cot {{200}^ \circ }}}}}$

Answer 1

The given question deals with basic simplification of trigonometric functions by using many simple trigonometric formulae such as $\sin (\pi + x) = - \sin x$ and $\cos (\pi + x) = - \cos x$ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We first convert all the trigonometric ratios into sine and cosine y=using some basic trigonometric formulae and identity and then simplify the expression.

Complete step by step answer:
In the given problem, we have to find the value of $\dfrac{{\dfrac{{2\sin {{140}^ \circ }\sec {{280}^ \circ }}}{{\sec {{220}^ \circ }}} + \dfrac{{\sec {{340}^ \circ }}}{{\cos ec{{20}^ \circ }}}}}{{\dfrac{{\cot {{200}^ \circ } - \tan {{280}^ \circ }}}{{\cot {{200}^ \circ }}}}}$ .
So, $\dfrac{{\dfrac{{2\sin {{140}^ \circ }\sec {{280}^ \circ }}}{{\sec {{220}^ \circ }}} + \dfrac{{\sec {{340}^ \circ }}}{{\cos ec{{20}^ \circ }}}}}{{\dfrac{{\cot {{200}^ \circ } - \tan {{200}^ \circ }}}{{\cot {{200}^ \circ }}}}} \\\$
Using $\cos ec(x) = \dfrac{1}{{\sin (x)}}$ and $\sec x = \dfrac{1}{{\cos x}}$, we get,
$\Rightarrow \dfrac{{\dfrac{{2\sin {{140}^ \circ }\cos {{220}^ \circ }}}{{\cos {{280}^ \circ }}} + \dfrac{{\sin {{20}^ \circ }}}{{\cos {{340}^ \circ }}}}}{{1 - \dfrac{{\tan {{280}^ \circ }}}{{\cot {{200}^ \circ }}}}} \\\$
Using $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$, we get,
$\Rightarrow \dfrac{{\dfrac{{2\sin {{140}^ \circ }\cos {{220}^ \circ }}}{{\cos {{280}^ \circ }}} + \dfrac{{\sin {{20}^ \circ }}}{{\cos {{340}^ \circ }}}}}{{1 - \dfrac{{\left( {\dfrac{{\sin {{280}^ \circ }}}{{\cos {{280}^ \circ }}}} \right)}}{{\left( {\dfrac{{\cos {{200}^ \circ }}}{{\sin {{200}^ \circ }}}} \right)}}}}$

Using the trigonometric formulae $\cos \left( {2\pi - \theta } \right) = \cos \theta$ and $\sin \left( {\pi - \theta } \right) = \sin \theta$, we get,
$\dfrac{{\dfrac{{2\sin {{40}^ \circ }\cos {{140}^ \circ }}}{{\cos {{80}^ \circ }}} + \dfrac{{\sin {{20}^ \circ }}}{{\cos {{20}^ \circ }}}}}{{1 - \left( {\dfrac{{\sin {{200}^ \circ }\sin {{280}^ \circ }}}{{\cos {{200}^ \circ }\cos {{280}^ \circ }}}} \right)}}$
Now, using the trigonometric identities $\cos \left( {\pi - \theta } \right) = - \cos \theta$ and $\sin 2x = 2\sin x\cos x$, we get,
$\dfrac{{\dfrac{{ - 2\sin {{40}^ \circ }\cos {{40}^ \circ }}}{{\cos {{80}^ \circ }}} + \dfrac{{\sin {{20}^ \circ }}}{{\cos {{20}^ \circ }}}}}{{\left( {\dfrac{{\cos {{200}^ \circ }\cos {{280}^ \circ } - \sin {{200}^ \circ }\sin {{280}^ \circ }}}{{\cos {{200}^ \circ }\cos {{280}^ \circ }}}} \right)}} \\\$
$\Rightarrow \dfrac{{\dfrac{{ - \sin {{80}^ \circ }}}{{\cos {{80}^ \circ }}} + \dfrac{{\sin {{20}^ \circ }}}{{\cos {{20}^ \circ }}}}}{{\left( {\dfrac{{\cos {{200}^ \circ }\cos {{280}^ \circ } - \sin {{200}^ \circ }\sin {{280}^ \circ }}}{{\cos {{200}^ \circ }\cos {{280}^ \circ }}}} \right)}}$

Now, using the compound angle formulae of sine and cosine $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$. So, we get,
$\Rightarrow \dfrac{{\dfrac{{ - \sin {{80}^ \circ }\cos {{20}^ \circ } + \cos {{80}^ \circ }\sin {{20}^ \circ }}}{{\cos {{80}^ \circ }\cos {{20}^ \circ }}}}}{{\left( {\dfrac{{\cos {{200}^ \circ }\cos {{280}^ \circ } - \sin {{200}^ \circ }\sin {{280}^ \circ }}}{{\cos {{200}^ \circ }\cos {{280}^ \circ }}}} \right)}}$
Rearranging the terms,
$\dfrac{{\dfrac{{\cos {{80}^ \circ }\sin {{20}^ \circ } - \sin {{80}^ \circ }\cos {{20}^ \circ }}}{{\cos {{80}^ \circ }\cos {{20}^ \circ }}}}}{{\left( {\dfrac{{\cos {{200}^ \circ }\cos {{280}^ \circ } - \sin {{200}^ \circ }\sin {{280}^ \circ }}}{{\cos {{200}^ \circ }\cos {{280}^ \circ }}}} \right)}} \\\$
$\Rightarrow \dfrac{{\dfrac{{\sin {{60}^ \circ }}}{{\cos {{80}^ \circ }\cos {{20}^ \circ }}}}}{{\dfrac{{\cos {{480}^ \circ }}}{{\cos {{200}^ \circ }\cos {{280}^ \circ }}}}}$
$\dfrac{{\sin {{60}^ \circ }\cos {{200}^ \circ }\cos {{280}^ \circ }}}{{\cos {{480}^ \circ }\cos {{80}^ \circ }\cos {{20}^ \circ }}}$

Now, we use trigonometric formulae $\cos \left( {\pi + \theta } \right) = - \cos \theta$ and $\cos \left( {2\pi + \theta } \right) = \cos \theta$,
$\Rightarrow \dfrac{{ - \sin {{60}^ \circ }\cos {{20}^ \circ }\cos {{80}^ \circ }}}{{\cos {{120}^ \circ }\cos {{80}^ \circ }\cos {{20}^ \circ }}} \\\$
Cancelling the common factors in numerator and denominator, we get,
$\dfrac{{ - \sin {{60}^ \circ }}}{{\cos {{120}^ \circ }}}$
Now, we know that $\cos \left( {\pi - \theta } \right) = - \cos \theta$
$\dfrac{{ - \sin {{60}^ \circ }}}{{ - \cos {{60}^ \circ }}}=\tan {60^ \circ }$
Now, we know the value of $\tan {60^ \circ }$. So, we get,
$\therefore \sqrt 3$

So, the value of $\dfrac{{\dfrac{{2\sin {{140}^ \circ }\sec {{280}^ \circ }}}{{\sec {{220}^ \circ }}} + \dfrac{{\sec {{340}^ \circ }}}{{\cos ec{{20}^ \circ }}}}}{{\dfrac{{\cot {{200}^ \circ } - \tan {{280}^ \circ }}}{{\cot {{200}^ \circ }}}}}$ is $\sqrt 3$.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: $\tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}}$ and $\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}}$ . Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers. We must remember the values for trigonometric expression of some standard angles such as ${60^ \circ }$, ${120^ \circ }$, etc. We must take care of calculations while solving such questions.