Question

How do you verify \[\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}} = {\left[ {\dfrac{{1 - \tan A}}{{1 - \cot A}}} \right]^2} = ({\tan ^2}A)\]?

Answer 1

Hint: Here in this question we have to prove the given inequality which is given in this question. This question involves the trigonometric function we should know about the trigonometry ratio. Hence by using the simple calculations we are going to prove the given inequality.

Complete step-by-step solution:
In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given inequality \[\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}} = ({\tan ^2}A)\]
Now we consider the LHS
 \[\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}\],
From the trigonometry identities we have \[1 + {\tan ^2}A = {\sec ^2}A\] and the \[1 + {\cot ^2}A = {\csc ^2}A\], now the given inequality is written as
\[ \Rightarrow \dfrac{{{{\sec }^2}A}}{{{{\csc }^2}A}}\]
The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent. From the reciprocal of the trigonometry ratios the above inequality can be written as
\[ \Rightarrow \dfrac{{\dfrac{1}{{{{\cos }^2}A}}}}{{\dfrac{1}{{{{\sin }^2}A}}}}\]
By taking the reciprocal this can be written as
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}A}}.\dfrac{{{{\sin }^2}A}}{1}\]
On multiplying this can be written as
\[ \Rightarrow \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}\]
Now by the property of trigonometry ratios this can be written as
\[ \Rightarrow {\tan ^2}A\]
\[ \Rightarrow RHS\]
Now we consider the other inequality
\[{\left[ {\dfrac{{1 - \tan A}}{{1 - \cot A}}} \right]^2} = ({\tan ^2}A)\]
We can \[\tan A = \dfrac{{\sin A}}{{\cos A}}\] and \[\cot A = \dfrac{{\cos A}}{{\sin A}}\]
\[ \Rightarrow {\left[ {\dfrac{{1 - \dfrac{{\sin A}}{{\cos A}}}}{{1 - \dfrac{{\cos A}}{{\sin A}}}}} \right]^2}\]
on simplifying we get
\[ \Rightarrow {\left[ {\dfrac{{\dfrac{{\cos A - \sin A}}{{\cos A}}}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}}} \right]^2}\]
Taking the reciprocal we get
\[ \Rightarrow {\left[ {\dfrac{{\cos A - \sin A}}{{\cos A}} \times \dfrac{{\sin A}}{{\sin A - \cos A}}} \right]^2}\]
Taking the minus in the second term
\[ \Rightarrow {\left[ {\dfrac{{\cos A - \sin A}}{{\cos A}} \times \dfrac{{\sin A}}{{ - \left( {\cos A - \sin A} \right)}}} \right]^2}\]
On simplifying we get
\[ \Rightarrow {\left[ {\dfrac{{ - \sin A}}{{\cos A}}} \right]^2}\]
We know that \[\tan A = \dfrac{{\sin A}}{{\cos A}}\]
\[ \Rightarrow {\left[ { - \tan A} \right]^2}\]
\[ \Rightarrow {\tan ^2}A\]
\[ \Rightarrow RHS\]
Here we have proved LHS = RHS.

Note: The question involves the trigonometric functions and we have to prove the trigonometric function. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities.