Question

Prove that: \[\sec A\left( 1-\sin A \right)\left( \operatorname{sec}A+\tan A \right)=1\]

Answer 1

Hint: In this question, we will write secA and tanA in terms of sinA and cosA and then use trigonometric identity to solve the question.

Complete step-by-step answer:
In a given question, we are asked to prove that
\[\sec A\left( 1-\sin A \right)\left( \sec A+\tan A \right)=1...........(i)\].
Now, we can write secant function and tangent function in terms of sine function and cosine function as given below,
$\begin{align}
  & \sec A=\dfrac{1}{\cos A}..............(ii) \\
 & \tan A=\dfrac{\sin A}{\cos A}.............(iii) \\
\end{align}$
Let us take left hand side of equation(i), we have,
$LHS=\sec A\left( 1-\sin A \right)\left( \sec A+\tan A \right)$
Substituting value of secA and tanA here with values of equation (i) and (ii), we get,
$\dfrac{1}{\cos A}\left( 1-\sin A \right)\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)$
Taking LCM and adding, we get,
$\dfrac{1}{\cos A}\left( 1-\sin A \right)\left( \dfrac{1+\sin A}{\cos A} \right)$
 Taking cosA common from denominator, we get,
$\dfrac{1}{{{\cos }^{2}}A}\left( 1-\sin A \right)\left( 1+\sin A \right)$
Applying distribution law, we get,
\[\dfrac{1}{{{\cos }^{2}}A}\left[ 1\left( 1+\sin A \right)-\sin A\left( 1+\sin A \right) \right]\]
Applying distributive law again, we get,
\[\dfrac{1}{{{\cos }^{2}}A}\left[ 1-{{\sin }^{2}}A \right]...........(iv)\]
Now, we know trigonometry identity of sine and cosine, which is
${{\sin }^{2}}A+{{\cos }^{2}}A=1$
Subtracting ${{\sin }^{2}}A$ from both sides of this equation, we get,
\[{{\cos }^{2}}A=1-{{\sin }^{2}}A\]
Using this value of \[1-{{\sin }^{2}}A\] in equation (iv), we get,
\[LHS=\dfrac{1}{{{\cos }^{2}}A}\left( {{\cos }^{2}}A \right)\]
Cancelling \[{{\cos }^{2}}A\] from numerator and denominator, we get,
LHS=1
Also, from equation (i), we have, the right hand side of equation is 1.
That is RHS=1
Therefore we get, LHS=RHS.
\[\Rightarrow \sec A\left( 1-\sin A \right)\left( \sec A+\tan A \right)=1\]
Hence, given equality is proved.

Note: In this type of question, try to write a given expression in such a form where you can use identities to simplify it and get desired results.