Question

Solve the following trigonometric equation and find the value
${\left( {\dfrac{{\sin {{47}^ \circ }}}{{\cos {{43}^ \circ }}}} \right)^2} + {\left( {\dfrac{{\cos {{43}^ \circ }}}{{\sin {{47}^ \circ }}}} \right)^2} - 4{\cos ^2}{45^ \circ }$

Answer 1

Hint: - Try to break the angle as a sum of other angles with multiple of${90^ \circ },{180^ \circ },{270^ \circ }\& {360^ \circ }$.
We have to find the value of${\left( {\dfrac{{\sin {{47}^ \circ }}}{{\cos {{43}^ \circ }}}} \right)^2} + {\left( {\dfrac{{\cos {{43}^ \circ }}}{{\sin {{47}^ \circ }}}} \right)^2} - 4{\cos ^2}{45^ \circ }$
As we know that
$\left[ {\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta ,\cos \left( {{{90}^ \circ } - \theta } \right) = \sin \theta \& \cos {{45}^ \circ } = \dfrac{1}{{\sqrt 2 }}} \right]$
So proceeding in the same way
$\Rightarrow {\left( {\dfrac{{\sin \left( {{{90}^ \circ } - {{43}^ \circ }} \right)}}{{\cos {{43}^ \circ }}}} \right)^2} + {\left( {\dfrac{{\cos \left( {{{90}^ \circ } - {{47}^ \circ }} \right)}}{{\sin {{47}^ \circ }}}} \right)^2} - 4{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} \\\ \Rightarrow 1 + 1 - 4\left( {\dfrac{1}{2}} \right) \\\ \Rightarrow 2 - 2 \\\ \Rightarrow 0 \\\$
Hence, the final value of the term is 0.
Note: - In case of some random degree angle in trigonometric don’t try to find out the value
of that term rather try to solve the problem by manipulation in degree by using different
trigonometric identities. One of them used has been mentioned above.