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ML Aggarwal Solutions Class 10 Mathematics Solutions for Trigonometric Identities Exercise 18 in Chapter 18 - Trigonometric Identities
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(cot A – 1)/ (2 – sec2 A) = cot A/ (1 + tan A)
First, we need to convert cotA and sec A in terms of sin A and cos A.
\begin{aligned} &\text { L. H.S. }=\frac{\cot A-1}{2-\sec ^{2} A}=\frac{\frac{\sin A}{\sin }-1}{2-\frac{1}{\cos ^{2} A}}\\ &=\frac{\frac{\cos A-\sin A}{\sin A}}{\frac{2 \cos ^{2} A-1}{\cos ^{2} A}}\\ &=\frac{\cos A-\sin A}{\sin A} \times \frac{\cos ^{2} A}{2 \cos ^{2} A-1}\\ &=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A\left(2 \cos ^{2} A-1\right)}\\ &=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A\left[2 \cos ^{2} A-\left(\sin ^{2} A+\cos ^{2} A\right)\right]}\\ &=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A\left[2 \cos ^{2} A-\sin ^{2} A-\cos ^{2} A\right]}\\ &=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A\left(\cos ^{2} A-\sin ^{2} A\right)}\\ &=\frac{\cos ^{2} A(\cos A-\sin A)}{\sin A(\cos A+\sin A)(\cos A-\sin A)}\\ &=\frac{\cos ^{2} A}{\sin A(\cos A+\sin A)} \end{aligned}
\begin{aligned} &\text { R.H.S. }=\frac{\cot A}{1+\tan A}=\frac{\frac{\cos A}{\sin A}}{1+\frac{\sin A}{\cos A}} \\ &=\frac{\frac{\cos A}{\sin A}}{\frac{\cos A+\sin A}{\cos A}} \\ &=\frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A+\sin A} \\ &=\frac{\cos 2 A}{\sin A(\cos A+\sin A)} \\ &\therefore \text { L.H.S. }=\text { R.H.S. } \end{aligned}
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