Question

Simplify the following equation
$\dfrac{{\cos e{c^2}67^\circ - {{\tan }^2}23^\circ }}{{{{\sec }^2}20^\circ - {{\cot }^2}70^\circ }}$

Answer 1

Hint: We need to know the basic trigonometric identities and formulae to solve this problem.
The given trigonometric expression is $\dfrac{{\cos e{c^2}67^\circ - {{\tan }^2}23^\circ }}{{{{\sec }^2}20^\circ - {{\cot }^2}70^\circ }}$
We have,
$$1 + {\cot ^2}\theta = \cos e{c^2}\theta $$
$1 + {\tan ^2}\theta = {\sec ^2}\theta $
Using these trigonometric identities, given function can be written as
$\dfrac{{\cos e{c^2}67^\circ - {{\tan }^2}23^\circ }}{{{{\sec }^2}20^\circ - {{\cot }^2}70^\circ }} = \dfrac{{1 + {{\cot }^2}67^\circ - {{\tan }^2}23^\circ }}{{1 + {{\tan }^2}20^\circ - {{\cot }^2}70^\circ }}$
We know that, $\tan (90 - \theta ) = \cot \theta $
$\cot \left( {90 - \theta } \right) = \tan \theta $
$ = \dfrac{{1 + {{\cot }^2}(90^\circ - 23^\circ ) - {{\tan }^2}23^\circ }}{{1 + {{\tan }^2}(90^\circ - 70^\circ ) - {{\cot }^2}70^\circ }}$
$ = \dfrac{{1 + {{\tan }^2}23^\circ - {{\tan }^2}23^\circ }}{{1 + {{\cot }^2}70^\circ - {{\cot }^2}70^\circ }}$
$ = \dfrac{1}{1} = 1$

Note:
$\cot \left( {90 - \theta } \right)\& \tan (90 - \theta )$are in the first quadrant. In the first quadrant all trigonometric functions are positive. So, tan and cot values in the first quadrant are positive.