Question

Find the value of $\dfrac{{\left( {\sin {{70}^ \circ } + \cos {{40}^ \circ }} \right)}}{{\left( {\cos {{70}^ \circ } + \sin {{40}^ \circ }} \right)}}$.
A. $1$
B. $\dfrac{1}{{\sqrt 3 }}$
C. $\sqrt 3$
D. $\dfrac{1}{2}$

Answer 1

The given problem requires us to simplify and find the value of the given trigonometric expression. The question requires thorough knowledge of trigonometric functions, formulae and identities. We should know the trigonometric identities $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$.
We must keep in mind the trigonometric identities while solving such questions.

Complete step by step answer:
In the given question, we are required to evaluate the value of $\dfrac{{\left( {\sin {{70}^ \circ } + \cos {{40}^ \circ }} \right)}}{{\left( {\cos {{70}^ \circ } + \sin {{40}^ \circ }} \right)}}$ using the basic concepts of trigonometry and identities.
So, we have, $\dfrac{{\left( {\sin {{70}^ \circ } + \cos {{40}^ \circ }} \right)}}{{\left( {\cos {{70}^ \circ } + \sin {{40}^ \circ }} \right)}}$.
We know that sine and cosine are complementary trigonometric functions. So, $\sin \left( x \right) = \cos \left( {{{90}^ \circ } - x} \right)$ and $\cos \left( x \right) = \sin \left( {{{90}^ \circ } - x} \right)$.
$\Rightarrow \dfrac{{\left( {\sin {{70}^ \circ } + \sin \left( {{{90}^ \circ } - {{40}^ \circ }} \right)} \right)}}{{\left( {\cos {{70}^ \circ } + \cos \left( {{{90}^ \circ } - {{40}^ \circ }} \right)} \right)}}$
Simplifying the angles of trigonometric angles further, we get,
$\Rightarrow \dfrac{{\sin {{70}^ \circ } + \sin {{50}^ \circ }}}{{\cos {{70}^ \circ } + \cos {{50}^ \circ }}}$

Now, we have the sum of sines in the numerator and the sum of cosines in the denominator.So, using the trigonometric identities $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ in the numerator and denominator respectively,we get,
$\Rightarrow \dfrac{{2\sin \left( {\dfrac{{{{70}^ \circ } + {{50}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{70}^ \circ } - {{50}^ \circ }}}{2}} \right)}}{{2\cos \left( {\dfrac{{{{70}^ \circ } + {{50}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{70}^ \circ } - {{50}^ \circ }}}{2}} \right)}}$
$\Rightarrow \dfrac{{2\sin \left( {{{60}^ \circ }} \right)\cos \left( {{{10}^ \circ }} \right)}}{{2\cos \left( {{{60}^ \circ }} \right)\cos \left( {{{10}^ \circ }} \right)}}$
Cancelling common factors in numerator and denominator, we get,
$\Rightarrow \dfrac{{\sin \left( {{{60}^ \circ }} \right)}}{{\cos \left( {{{60}^ \circ }} \right)}}$
Now, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$. So, we get,
$\Rightarrow \tan \left( {{{60}^ \circ }} \right)$
We know the value of $\tan {60^ \circ }$ as $\sqrt 3$. So, we get,
$\Rightarrow \sqrt 3$
So, we get the value of trigonometric expression $\dfrac{{\left( {\sin {{70}^ \circ } + \cos {{40}^ \circ }} \right)}}{{\left( {\cos {{70}^ \circ } + \sin {{40}^ \circ }} \right)}}$ as $\sqrt 3$.

Hence, the correct answer is option C.

Note: We must have thorough knowledge of trigonometric identities and formulae such as $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ to tackle these kind of problems. These identities are of vital importance for solving any question involving trigonometric functions and identities. All the trigonometric ratios can be converted into each other using the simple trigonometric identities. Algebraic operations and rules like transposition rules come into significant use while solving such problems.