Question

What is the value of $\left( {cosecA - sinA} \right)\left( {secA - cosA} \right)(tanA + cotA)$?
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint : : In trigonometry, trigonometric ratios give the values for the sides and angles in a triangle. It is given by using some ratios of the sides of the triangle with respect to its acute angles. Various trigonometric ratios are interrelated. The trigonometric ratios sin $A,\tan A,\cos A$ of an angle $A$ are very closely connected by a relation.
Formula used: it is clear from the definitions of the trigonometric ratios the for an acute angle A, we have
$\cos ecA = \dfrac{1}{{\sin A}},\cot A = \dfrac{1}{{\tan A}} \\\ \sec A = \dfrac{1}{{\cos A}} \;$

Complete step-by-step answer :
If any one of them is known the other two can be easily calculated by
( $\tan A = \dfrac{{\sin A}}{{\cos A}}$ )
Solving the given question let us substitute the values from the formulas and expanding we get,
$\left( {\dfrac{1}{{\sin A}} - sinA} \right)\left( {\dfrac{1}{{\cos A}} - cosA} \right)(tanA + \dfrac{1}{{\tan A}})$
Simplifying by taking the denominators as LCM.
$\left( {\dfrac{{1 - si{n^2}A}}{{\sin A}}} \right)\left( {\dfrac{{1 - co{s^2}A}}{{\cos A}}} \right)(\dfrac{{1 + ta{n^2}A}}{{\tan A}})$
Let us convert the equation in all sinA and cosA by the formula
$\left( {\dfrac{{1 - si{n^2}A}}{{\sin A}}} \right)\left( {\dfrac{{1 - co{s^2}A}}{{\cos A}}} \right)(\dfrac{{{{\sin }^2}A + {{\cos }^2}A}}{{\sin A\cos A}})$
We know that $\therefore {\sin ^2}A + {\cos ^2}A = 1$ therefore,
$\left( {\dfrac{{co{s^2}A}}{{\sin A}}} \right)\left( {\dfrac{{si{n^2}A}}{{\cos A}}} \right)(\dfrac{1}{{\sin A\cos A}})$
Now let us cancel out the trigonometric ratios common in numerator and denominator to eliminate the terms in the denominator,
$\left( {\dfrac{{cosA}}{{}}} \right)\left( {\dfrac{{sinA}}{{}}} \right)(\dfrac{1}{{\sin A\cos A}})$
Further repeating the process we observe that the terms left in both numerator and denominator are common therefore it will cancel each other out and we will be left with the remainder 1.
Therefore $\left( {cosecA - sinA} \right)\left( {secA - cosA} \right)(tanA + cotA)$=1 and the correct option is B.
So, the correct answer is “Option B”.

Note : It must be observed that ${\sin ^2}A$ is a square of the sine of angle A and similarly ${\cos ^2}A$ is the square of the cosine of the angle A which are totally different from $(sinA^2)$ and $(cosA^2)$.