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

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

Explanation

Solution

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(1−sin⁡A)(sec⁡A+tan⁡A)=1...........(i)\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,
sec⁡A=1cos⁡A..............(ii) tan⁡A=sin⁡Acos⁡A.............(iii) \begin{aligned} & \sec A=\dfrac{1}{\cos A}..............(ii) \\\ & \tan A=\dfrac{\sin A}{\cos A}.............(iii) \\\ \end{aligned}
Let us take left hand side of equation(i), we have,
LHS=sec⁡A(1−sin⁡A)(sec⁡A+tan⁡A)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,
1cos⁡A(1−sin⁡A)(1cos⁡A+sin⁡Acos⁡A)\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,
1cos⁡A(1−sin⁡A)(1+sin⁡Acos⁡A)\dfrac{1}{\cos A}\left( 1-\sin A \right)\left( \dfrac{1+\sin A}{\cos A} \right)
Taking cosA common from denominator, we get,
1cos⁡2A(1−sin⁡A)(1+sin⁡A)\dfrac{1}{{{\cos }^{2}}A}\left( 1-\sin A \right)\left( 1+\sin A \right)
Applying distribution law, we get,
1cos⁡2A[1(1+sin⁡A)−sin⁡A(1+sin⁡A)]\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,
1cos⁡2A[1−sin⁡2A]...........(iv)\dfrac{1}{{{\cos }^{2}}A}\left[ 1-{{\sin }^{2}}A \right]...........(iv)
Now, we know trigonometry identity of sine and cosine, which is
sin⁡2A+cos⁡2A=1{{\sin }^{2}}A+{{\cos }^{2}}A=1
Subtracting sin⁡2A{{\sin }^{2}}A from both sides of this equation, we get,
cos⁡2A=1−sin⁡2A{{\cos }^{2}}A=1-{{\sin }^{2}}A
Using this value of 1−sin⁡2A1-{{\sin }^{2}}A in equation (iv), we get,
LHS=1cos⁡2A(cos⁡2A)LHS=\dfrac{1}{{{\cos }^{2}}A}\left( {{\cos }^{2}}A \right)
Cancelling cos⁡2A{{\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.
⇒sec⁡A(1−sin⁡A)(sec⁡A+tan⁡A)=1\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.