Question

Prove the given trigonometric expression \[\left( 1-co{{s}^{2}}A \right)cose{{c}^{2}}A=1\].

Answer 1

Hint: We have to prove the expression \[\left( 1-co{{s}^{2}}A \right)cose{{c}^{2}}A=1\] . We know the identity that, \[{{\sin }^{2}}A+co{{s}^{2}}A=1\] . Now from this identity put the value of \[(1-co{{s}^{2}}A)\] in the expression \[\left( 1-co{{s}^{2}}A \right)cose{{c}^{2}}A\] . We know that \[\operatorname{cosecA}\] is reciprocal of \[sinA\] . That is, \[\operatorname{cosecA}=\dfrac{1}{sinA}\] . Now, using this transform the expression and solve further.

Complete step-by-step solution -
According to the question, we have to prove \[\left( 1-co{{s}^{2}}A \right)cose{{c}^{2}}A=1\] .
First of all, we need to simplify the given expression, \[\left( 1-co{{s}^{2}}A \right)cose{{c}^{2}}A\] …………..(1)
We know the identity, \[{{\sin }^{2}}A+co{{s}^{2}}A=1\] .
Now from this identity, we can find the value of \[(1-co{{s}^{2}}A)\] .
Solving the identity we get,
\[\begin{align}
  & {{\sin }^{2}}A+co{{s}^{2}}A=1 \\
 & \Rightarrow {{\sin }^{2}}A=1-co{{s}^{2}}A \\
\end{align}\]
Now, we have \[(1-co{{s}^{2}}A)=si{{n}^{2}}A\] …………….(2)
Now, from equation (1) and equation (2), we get
\[\left( 1-co{{s}^{2}}A \right)cose{{c}^{2}}A\]
\[=si{{n}^{2}}A.cose{{c}^{2}}A\] …………………..(3)
We know that \[\operatorname{cosecA}\] is reciprocal of \[sinA\] . That is, \[\operatorname{cosecA}=\dfrac{1}{sinA}\] ………….(4)
Now, from equation (3) and equation (4), we get
\[\begin{align}
  & =si{{n}^{2}}A.cose{{c}^{2}}A \\
 & =si{{n}^{2}}A.\dfrac{1}{si{{n}^{2}}A} \\
 & =1 \\
\end{align}\]
Solving the LHS, we got 1 and in RHS, we also have 1.
So, LHS = RHS.
Hence, proved.

Note: We can also solve this question in another way.
We know that \[\operatorname{cosecA}\] is reciprocal of \[sinA\] . That is, \[\operatorname{cosecA}=\dfrac{1}{sinA}\]………..(1)
Using equation (1), we can transform the given expression \[\left( 1-co{{s}^{2}}A \right)cose{{c}^{2}}A\] .
\[\left( 1-co{{s}^{2}}A \right)cose{{c}^{2}}A\]
\[=\left( 1-co{{s}^{2}}A \right)\dfrac{1}{si{{n}^{2}}A}\] ………………..(2)
We know the identity, \[{{\sin }^{2}}A+co{{s}^{2}}A=1\] .
On solving we get
\[{{\sin }^{2}}A=1-co{{s}^{2}}A\] ………………..(3)
From equation (2) and equation (3), we get
\[=\left( 1-co{{s}^{2}}A \right)\dfrac{1}{si{{n}^{2}}A}\]
\[\begin{align}
  & =\left( 1-co{{s}^{2}}A \right)\dfrac{1}{(1-{{\cos }^{2}}A)} \\
 & =1 \\
\end{align}\]
Solving the LHS, we got 1 and in RHS, we also have 1.
So, LHS = RHS.
Hence, proved.