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

Last updated date: 15th Sep 2024
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Hint: 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.

\begin{align} & \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{align}
\begin{align} &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{align}
(Using the identity of trigonometry which is $({{\sin }^{2}}A+{{\cos }^{2}}A=1)$ )