What is the value of \[\left( {cosecA - sinA} \right)\left( {secA - cosA} \right)(tanA + cotA)\]?
A. 0
B. 1
C. 2
D. 3
Answer
644.1k+ views
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) $.
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) $.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

