Question

The value of $\sqrt {\dfrac{{1 + \cos 2\theta }}{{1 - \cos 2\theta }}}$ is not equal to
(1) $\csc 2\theta - \cot 2\theta$
(2) $\dfrac{{1 + \cos 2\theta }}{{\sin \theta }}$
(3) $\csc 2\theta + \cot 2\theta$
(4) $\cot 2\theta \left( {\sec 2\theta + 1} \right)$

Answer 1

Hint : In order to evaluate the given question , we must first know the trigonometric ratios and the trigonometric identities. Every trigonometric function and formulae are designed on the basis of three primary ratios. Sine, Cosine and tangents are these ratios in trigonometry based on Perpendicular, Hypotenuse and Base of a right triangle . In order to simplify the above equation containing sin , cos and cot functions . Also , we should know the reciprocals of the ratios .For this particular question we need to know the Pythagorean identities through which on applying we can get our required solution .

Complete step by step solution:
In the above trigonometric question, we need to solve and simplify the trigonometric equation by using some of the trigonometric identities and the reciprocals.
For evaluating the given question $\sqrt {\dfrac{{1 + \cos 2\theta }}{{1 - \cos 2\theta }}}$ , we must recall the trigonometric identity related to this given question.
We are going to multiply the numerator and the denominator both by $(1 + \cos \theta )$ . So that we can apply the identity as stated below -
We can apply the formula or we can say identity $(a + b)(a - b) = {a^2} - {b^2}$
Which we can apply on the given equation $\sqrt {\dfrac{{1 + \cos 2\theta }}{{1 - \cos 2\theta }}}$
We can substitute 1 on the place of a and cos on the place of b , we get –
$= [1 - \cos A][1 + \cos A] = {1^2} - {\cos ^2}A = 1 - {\cos ^2}A$
The final equation we get = $\sqrt {\dfrac{{{{\left( {1 + \cos 2\theta } \right)}^2}}}{{\left( {{1^2} - {{\cos }^2}2\theta } \right)}}}$
Here we are going to use the Pythagorean identities through which on applying we can
Solve are as follows
we use the Pythagorean identity as follows –
${\sin ^2}A + {\cos ^2}A = 1 \\\ 1 - {\cos ^2}A = {\sin ^2}A \;$
Substituting this in our final equation, we get –
$\sqrt {\dfrac{{{{\left( {1 + \cos 2\theta } \right)}^2}}}{{\left( {si{n^2}2\theta } \right)}}}$
Further , we can get rid of radical sign as we have squares under the radical .
$\dfrac{{\left( {1 + \cos 2\theta } \right)}}{{\left( {sin2\theta } \right)}}$
Simplifying further , we can split the denominator with the different terms of the numerator .
$\dfrac{1}{{sin2\theta }} + \dfrac{{\cos 2\theta }}{{sin2\theta }}$
We can substitute the ${\cot ^2}2\theta$ in the place of $\dfrac{{\cos 2\theta }}{{sin2\theta }}$ in the equation because , we get –
$\dfrac{1}{{sin2\theta }} + \cot 2\theta$
Again , using the reciprocal identity we know that the cosec is the reciprocal of sin ,
$\sin A = \dfrac{1}{{\cos ecA}}$
We will substitute this in our final equation , we get –
$\Rightarrow \cos ec2\theta + \cot 2\theta$
So, the required simplification while solving which can be equal to the given question
$\sqrt {\dfrac{{1 + \cos 2\theta }}{{1 - \cos 2\theta }}}$ comes out to be as follows
$\cos ec2\theta + \cot 2\theta$
$\Rightarrow \dfrac{{\left( {1 + \cos 2\theta } \right)}}{{\left( {sin2\theta } \right)}}$
and the correct option is ( 1 ) and ( 4 ) .
If we have multiplied by $(1 - \cos \theta )$ by both numerator and the denominator in the starting as step 1 , we would have get the solution as $\cos ec2\theta - \cot 2\theta$ , but for now this is not equal to the given question .
Also , $\cot 2\theta (\sec 2\theta + 1)$ can be solved as $1 + \cot 2\theta$ , so this is also not equal to the given question
Therefore, the correct option is (1) and ( 4 ) .
So, the correct answer is “(1) and ( 4 )”.

Note : Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus .
Remember the Pythagorean and Reciprocal identities as per the application and usage .
We should know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta$ .
As a matter of fact, $\sin \theta$ and $\tan \theta$ and their reciprocals, $\cos ec\theta$ and $\cot \theta$ are odd functions whereas $\cos \theta$ and its reciprocal $\sec \theta$ are even functions .
One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.