Question

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
$\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$

Answer 1

To do this question, firstly we will write $\sec x=\dfrac{1}{\cos x}$ on the left hand side. Then, we will take LCM on the left hand side and then we will simplify the numerator of fraction on the left hand side. After simplifying, we will multiply numerator and denominator by 1 – cosA for more simplification. Then, we will use the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to simplify the numerator. After that, we will take the common factors out from both numerator and denominator and at last we will leave with only expression equals to RHS.

Complete step by step answer:
In such questions, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.
The below mentioned formulae may be used before solving, in the solution which is as follows

& \tan x=\dfrac{\sin x}{\cos x} \\\ & \cot x=\dfrac{\cos x}{\sin x} \\\ & \cos ecx=\dfrac{1}{\sin x} \\\ & \sec x=\dfrac{1}{\cos x} \\\ \end{aligned}$$ Now, these are the results that would be used to prove the proof mentioned in this question as using these identities, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other. In this particular question, we will first convert all the trigonometric functions in terms of sin and cos function and then we can convert the expression in terms of tan and cot function and then we will try to make the L.H.S. and the R.H.S. equal. As mentioned in the question, we have to prove the given expression. Now, we will start with the left hand side that is L.H.S. and try to make the necessary changes that are given in the hint, first, as follows $$\begin{aligned} &LHS;= \dfrac{1+\sec A}{\sec A} \\\ & \Rightarrow LHS =\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}} \\\ & \Rightarrow LHS =\dfrac{\left( \cos A+1 \right)\left( 1-\cos A \right)}{\left( 1-\cos A \right)} \\\ & \Rightarrow LHS =\dfrac{1-{{\cos }^{2}}A}{\left( 1-\cos A \right)} \\\ & ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1) \\\ & \Rightarrow LHS =\dfrac{{{\sin }^{2}}A}{1-\cos A} \\\ \end{aligned}$$ (Using the identity of trigonometry which is $$({{\sin }^{2}}A+{{\cos }^{2}}A=1)$$ ) Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved. **Note:** In these types of questions, order of each and every step matters a lot, so when you solve the question try to simplify the equation first and then use the trigonometric formulas, as this will help in getting the final answer in less steps. Try not to make any calculation errors while doing the solution of the question.