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

Last updated date: 22nd Jul 2024
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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.