Question

Prove the following trigonometric expression:
\[\operatorname{cosec}\left( 90+\theta \right)=\sec \theta \]

Answer 1

Hint: First of all take, LHS of the given equation. Now use, \[\operatorname{cosec}x=\dfrac{1}{\sin x}\]. Now use the formula of sin (A + B) = sin A cos B + cos A sin B and substitute the value of \[\sin {{90}^{o}}\] from the trigonometric table. Now use, \[\dfrac{1}{\cos x}=\sec x\] to get the required value.

Complete step-by-step solution -
In this question, we have to prove that
\[\operatorname{cosec}\left( 90+\theta \right)=\sec \theta \]
Let us consider the LHS of the equation given in the question.
\[LHS=\operatorname{cosec}\left( 90+\theta \right)\]
We know that \[\operatorname{cosec}x=\dfrac{1}{\sin x}\]. By using this in the above equation, we get,
\[LHS=\dfrac{1}{\sin \left( 90+\theta \right)}\]
We know that sin (A + B) = sin A cos B + cos A sin B. By using this in the above equation, we get,
\[LHS=\dfrac{1}{\sin 90\cos \theta +\cos 90\sin \theta }....\left( i \right)\]
Let us find the value of sin 90 and cos 90 from the trigonometric table for general angles.

	\[\sin \theta \]	\[\cos \theta \]	\[\tan \theta \]	\[\operatorname{cosec}\theta \]	\[\sec \theta \]	\[\cot \theta \]
0	0	1	0	-	1	-
\[\dfrac{\pi }{6}\]	\[\dfrac{1}{2}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{3}}\]	2	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{3}\]
\[\dfrac{\pi }{4}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{\sqrt{2}}\]	1	\[\sqrt{2}\]	\[\sqrt{2}\]	1
\[\dfrac{\pi }{3}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{2}\]	\[\sqrt{3}\]	\[\dfrac{2}{\sqrt{3}}\]	2	\[\dfrac{1}{\sqrt{3}}\]
\[\dfrac{\pi }{2}\]	1	0	-	1	-	0

From the above table, we get sin 90 = 1 and cos 90 = 0. By substituting these in equation (i), we get
\[LHS=\dfrac{1}{1\left( \cos \theta \right)+0\left( \sin \theta \right)}\]
So, we get,
\[LHS=\dfrac{1}{\cos \theta +0}\]
\[LHS=\dfrac{1}{\cos \theta }\]
We know that \[\dfrac{1}{\cos x}=\sec x\]. By using this in the above equation, we get,
\[LHS=\sec \theta \]
LHS = RHS
Hence proved.
So, we have proved that \[\operatorname{cosec}\left( 90+\theta \right)=\sec \theta \]

Note: Students can also use these equations of complementary angles as direct formulas in the various questions. Students just need to remember that \[\sin \left( 90-\theta \right)=\cos \theta ,\cos \left( 90-\theta \right)=\sin \theta ,\sin \left( 90+\theta \right)=\cos \theta \text{ and }\cos \left( 90+\theta \right)=-\sin \theta \] and from this, they can derive all the other formulas by using \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta =\dfrac{1}{\cot \theta },\sin \theta =\dfrac{1}{\operatorname{cosec}\theta }\text{ and }\cos \theta =\dfrac{1}{\sec \theta }\]. Students need to remember the formulas of trigonometry to save from lengthy and complicated solutions.