Question

How do you prove this trigonometric expression show your work on paper

Answer 1

Solution!!

$\sf \dfrac{\sec \theta }{\cos \theta } - \dfrac{\tan \theta }{\cot \theta }=1$

Taking LHS

$\sf =\dfrac{\sec \theta }{\cos \theta } - \dfrac{\tan \theta }{\cot \theta }$

Use the following relations to transform the expressions:

→ $\sf \sec \theta = \dfrac{1}{\cos \theta }$

→ $\sf \tan \theta = \dfrac{\sin \theta }{\cos \theta }$

→ $\sf \cot \theta = \dfrac{\cos \theta }{\sin \theta }$

$\sf =\dfrac{\frac{1}{\cos \theta }}{\cos \theta }-\dfrac{\frac{\sin \theta }{\cos \theta }}{\frac{\cos \theta }{\sin \theta }}$

Simplify the complex fractions.

$\sf =\dfrac{1}{\cos ^{2}\theta }-\dfrac{\sin ^{2}\theta }{\cos ^{2}\theta }$

Write the numerators above a common denominator.

$\sf =\dfrac{1-\sin ^{2}\theta }{\cos ^{2}\theta }$

Simplify the expression using:

→ $\sf 1-\sin ^{2}\theta =\cos ^{2}\theta$

$\sf =\dfrac{\cos ^{2}\theta }{\cos^{2}\theta }$

Any expression divided by itself equals 1.

= 1

LHS = RHS

Hence, proved.