# Simplify the following equation

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

Answer

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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.

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.

Last updated date: 29th Sep 2023

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