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Question: If \(3\cot A = 4\), check whether \(\dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\...

If 3cotA=43\cot A = 4, check whether 1tan2A1+tan2A=cos2Asin2A\dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A or not?

Explanation

Solution

With the help of the condition given i.e. 3cotA=43\cot A = 4, figure out the trigonometric values which are needed and substitute In 1tan2A1+tan2A=cos2Asin2A\dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A, and check whether it satisfies or not.

Complete step-by-step answer:
Given that 3cotA=43\cot A = 4, Draw a right angled triangle and find the value needed, with the help of the values, figure out the trigonometric values needed.

We know that cotA=AdjacentsideOppositeside=43\cot A = \dfrac{{Adjacent\,side}}{{Opposite\,side}} = \dfrac{4}{3}
Take 1tan2A1+tan2A\dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}and substitute the value of tanA\tan A, where tanA=34\tan A = \dfrac{3}{4}
1tan2A1+tan2A 1(34)21+(34)2 1(916)1+(916) (16916)(16+916) 725.......(1)  \Rightarrow \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} \\\ \Rightarrow \dfrac{{1 - {{\left( {\dfrac{3}{4}} \right)}^2}}}{{1 + {{\left( {\dfrac{3}{4}} \right)}^2}}} \\\ \Rightarrow \dfrac{{1 - \left( {\dfrac{9}{{16}}} \right)}}{{1 + \left( {\dfrac{9}{{16}}} \right)}} \\\ \Rightarrow \dfrac{{\left( {\dfrac{{16 - 9}}{{16}}} \right)}}{{\left( {\dfrac{{16 + 9}}{{16}}} \right)}} \\\ \Rightarrow \dfrac{7}{{25}}.......\left( 1 \right) \\\
Now take cos2Asin2A{\cos ^2}A - {\sin ^2}Aand substitute the value of cosA\cos A and sinA\sin A, where cosA=AdjacentsideHypotenuseside=45\cos A = \dfrac{{Adjacent\,side}}{{{\text{Hypotenuse}}\,{\text{side}}}} = \dfrac{4}{5}and sinA=OppositeHypotenuse=35\sin A = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}} = \dfrac{3}{5}
cos2Asin2A (45)2(35)2 1625925 725.......(2)  \Rightarrow {\cos ^2}A - {\sin ^2}A \\\ \Rightarrow {\left( {\dfrac{4}{5}} \right)^2} - {\left( {\dfrac{3}{5}} \right)^2} \\\ \Rightarrow \dfrac{{16}}{{25}} - \dfrac{9}{{25}} \\\ \Rightarrow \dfrac{7}{{25}}.......\left( 2 \right) \\\
So, here from (1) and (2), both satisfy the equation. Hence, 1tan2A1+tan2A=cos2Asin2A\dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A.

Note: While solving trigonometric problems, know the values of trigonometric functions, it becomes easy while solving trigonometric problems. Try to draw the diagram while solving trigonometric problems.