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Question

Question: How do you verify \( \cos ec2x = \dfrac{{\cos ecx}}{{2\cos x}} \)...

How do you verify cosec2x=cosecx2cosx\cos ec2x = \dfrac{{\cos ecx}}{{2\cos x}}

Explanation

Solution

Hint : In the given question, we have to verify that cosec2x=cosecx2cosx\cos ec2x = \dfrac{{\cos ecx}}{{2\cos x}} that is we have to show that the part on the left side of the equal to sign is equal to the part on the right side of the equal to sign. In simple words, we have to prove that the left-hand side of the given equation is equal to the right-hand side. For that, we will take any one side and solve it using the trigonometric ratios or identities and make it equal to the other side.

Complete step-by-step answer :
Solving left-hand side –
We know that
cosecx=1sinx cosec2x=1sin2x  \cos ecx = \dfrac{1}{{\sin x}} \\\ \Rightarrow \cos ec2x = \dfrac{1}{{\sin 2x}} \\\
Now, we know that sin2x=2sinxcosx\sin 2x = 2\sin x\cos x , so –
1sin2x=12sinxcosx\dfrac{1}{{\sin 2x}} = \dfrac{1}{{2\sin x\cos x}}
As we know 1sinx=cosecx\dfrac{1}{{\sin x}} = \cos ecx ,
cosec2x=cosecx2cosx\Rightarrow \cos ec2x = \dfrac{{\cos ecx}}{{2\cos x}}
We see that the left-hand side has become equal to the right-hand side.
Hence, we have verified that cosec2x=cosecx2cosx\cos ec2x = \dfrac{{\cos ecx}}{{2\cos x}} .
So, the correct answer is “ cosec2x=cosecx2cosx\cos ec2x = \dfrac{{\cos ecx}}{{2\cos x}} ”.

Note : The trigonometric ratios are the ratio of two sides of a right-angled triangle. The sine function is the ratio of perpendicular and the hypotenuse of the right-angled triangle, cos function is the ratio of the base and the hypotenuse of the right-angled triangle and tan function is the ratio of the perpendicular and the base of the right-angled triangle. Cosecant, secant and cotangent functions are the reciprocal of the sine, cos and tangent function respectively, thus all the trigonometric functions are interrelated with each other and one can be converted into another using this knowledge or the trigonometric identities like we have used the identity sin2x=2sinxcosx\sin 2x = 2\sin x\cos x in the given question.