Question

Prove that $\left( {1 + \dfrac{1}{{{{\tan }^2}A}}} \right)\left( {1 + \dfrac{1}{{{{\cot }^2}A}}} \right) = \dfrac{1}{{({{\sin }^2}A - {{\sin }^4}A)}}$

Answer 1

Hint : To solve these types of questions. We need to remember all the trigonometry identities. So that it will give you an idea how to proceed. We can see that we have ${\tan ^2}A$ which can be written as $\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}$. Similarly for the${\cot ^2}A$, which is a reciprocal of ${\tan ^2}A$. Substituting these in the above and on simplifying we can use Pythagoras trigonometric identity ${\sin ^2}A + {\cos ^2}A = 1$.

Complete step-by-step answer :
Now given, $\left( {1 + \dfrac{1}{{{{\tan }^2}A}}} \right)\left( {1 + \dfrac{1}{{{{\cot }^2}A}}} \right) = \dfrac{1}{{({{\sin }^2}A - {{\sin }^4}A)}}$
Take $L.H.S = \left( {1 + \dfrac{1}{{{{\tan }^2}A}}} \right)\left( {1 + \dfrac{1}{{{{\cot }^2}A}}} \right)$ and $R.H.S. = \dfrac{1}{{({{\sin }^2}A - {{\sin }^4}A)}}$
Now take $L.H.S = \left( {1 + \dfrac{1}{{{{\tan }^2}A}}} \right)\left( {1 + \dfrac{1}{{{{\cot }^2}A}}} \right)$
Taking L.C.M in the both the bracket’s and solving separately we get,
$L.H.S = \left( {\dfrac{{{{\tan }^2}A + 1}}{{{{\tan }^2}A}}} \right)\left( {\dfrac{{{{\cot }^2}A + 1}}{{{{\cot }^2}A}}} \right)$
We know the trigonometric identity ${\tan ^2}A + 1 = {\sec ^2}A$ and ${\cot ^2}A + 1 = {\csc ^2}A$
On substituting we get ,
$\Rightarrow L.H.S = \left( {\dfrac{{{{\sec }^2}A}}{{{{\tan }^2}A}}} \right)\left( {\dfrac{{{{\csc }^2}A}}{{{{\cot }^2}A}}} \right)$
$\Rightarrow L.H.S = \left( {\dfrac{{{{\sec }^2}A.{{\csc }^2}A}}{{{{\tan }^2}A.{{\cot }^2}A}}} \right)$
We know that tangent and the cotangent are reciprocals and hence in the denominator of the above equation it gets canceled out.
We obtain, $\Rightarrow L.H.S = {\sec ^2}A{\csc ^2}A$
Again we know that sine is a reciprocal of cosecant and cosine is a reciprocal of secant.
Then,
$\Rightarrow L.H.S = \dfrac{1}{{({{\sin }^2}A)({{\cos }^2}A)}}$
In the denominator of the above equation we have cosine function, we convert into sine function.
We know that ${\sin ^2}A + {\cos ^2}A = 1$, we can rewrite it as ${\cos ^2}A = 1 - {\sin ^2}A$. So that in the denominator we will have only sine function.
We get,
$\Rightarrow L.H.S. = \dfrac{1}{{{{\sin }^2}A(1 - {{\sin }^2}A)}}$
Removing brackets by multiplying in the denominator we get,
$\Rightarrow L.H.S. = \dfrac{1}{{({{\sin }^2}A - {{\sin }^4}A)}}$, which is $R.H.S.$
So we obtain,
$\Rightarrow L.H.S. = R.H.S$
Thus we have proved $\left( {1 + \dfrac{1}{{{{\tan }^2}A}}} \right)\left( {1 + \dfrac{1}{{{{\cot }^2}A}}} \right) = \dfrac{1}{{({{\sin }^2}A - {{\sin }^4}A)}}$
Hence, the result.

Note : In above we simply used the reciprocal identities which we know. Note that for a given angle $A$ each ratio stays the same no matter how big or small the triangle is. We can also convert sine into cosine and cosine into sine function by using the Pythagoras trigonometric identity ${\sin ^2}A + {\cos ^2}A = 1$, which we did in the above method. Always remember basic trigonometric identities.