Question

Solve $[1 - \cos A][1 + \cos A][1 + {\cot ^2}A] = ?$

Answer 1

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 $[1 - \cos A][1 + \cos A][1 + {\cot ^2}A] = ?$ , we must recall the trigonometric identity related to this given question .
Here we are going to use the Pythagorean identities through which on applying we can
Solve are as follows
$\Rightarrow 1 + {\cot ^2}\theta = {\csc ^2}\theta$
We can substitute the ${\csc ^2}\theta$in the place of $1 + {\cot ^2}\theta$in the equation , we get –

$$\Rightarrow [1 - \cos A][1 + \cos A][1 + {\cot ^2}A] \\\ \Rightarrow [1 - \cos A][1 + \cos A][{\csc ^2}A] \\\$$

Thereafter , 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 $= [1 - \cos A][1 + \cos A][{\csc ^2}A]$
We can substitute 1 on the place of a and cos on the place of b , we get –
$\Rightarrow [1 - \cos A][1 + \cos A] = {1^2} - {\cos ^2}A = 1 - {\cos ^2}A$
The final equation we get = $= [1 - {\cos ^2}A][{\csc ^2}A]$
Again , we use the Pythagorean identity as follows –
$\Rightarrow {\sin ^2}A + {\cos ^2}A = 1 \\\ \Rightarrow 1 - {\cos ^2}A = {\sin ^2}A \\\$
Substituting this in our final equation , we get –
$\Rightarrow [{\sin ^2}A][{\csc ^2}A]$
Again , using the reciprocal identity we know that the cosec is the reciprocal of sin ,
$\Rightarrow \sin A = \dfrac{1}{{\cos ecA}}$
We will substitute this in our final equation , we get –
$\Rightarrow [{\sin ^2}A][{\csc ^2}A]$
$\Rightarrow [{\sin ^2}A] \times \left[ {\dfrac{1}{{{{\sin }^2}A}}} \right]$
$\Rightarrow 1$

Hence the correct answer is ‘1’.

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 .