Question

Prove that:

(i) (sin5A - sin3A) ÷ (cos5A + cos3A) = tanA

(ii) (sinA + sin3A) ÷ (cosA + cos3A) = tan2A

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Answer 1

Step-by-step explanation:

Refer to the attachment!!

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Answer 2

${\mathcal{\red{SOLUTIONS}}}$

$\tt{1).\;\dfrac{\sin5A -\sin3A}{\cos5A+\cos3A} = \tan A}$

We take LHS,

$\tt{\implies \dfrac{\sin5A -\sin3A}{\cos5A+\cos3A}}$

$\tt{\implies \dfrac{2\cos\bigg(\dfrac{5A+3A}{2}\bigg)\;\sin\bigg(\dfrac{5A-3A}{2}\bigg)}{2\cos\bigg(\dfrac{5A+3A}{2}\bigg)\;\cos\bigg(\dfrac{5A-3A}{2}\bigg)}}$

$\tt{\implies \dfrac{2\cos\bigg(\dfrac{8A}{2}\bigg)\;\sin\bigg(\dfrac{2A}{2}\bigg)}{2\cos \bigg(\dfrac{8A}{2}\bigg)\;\cos\bigg(\dfrac{2A}{2}\bigg)}}$

$\tt{\implies \dfrac{2\cos(4A)\;\sin(A)}{2\cos(4A)\;\cos(A)}}$

Now numerator have $\bf{2\cos 4A}$ common so they will get cancelled and hence we will get,

$\tt{\implies \dfrac{\sin(A)}{\cos(A)} = \tan A}$

Hence Proved

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$\tt{2).\; \dfrac{\sin A + \sin 3A}{\cos A + \cos 3A} = \tan 2A}$

We take LHS,

$\tt{\implies \dfrac{\sin A + \sin 3A}{\cos A + \cos 3A}}$

$\tt{\implies \dfrac{\sin A + 3\sin A- 4\sin^{3}A}{\cos A + 4\cos^{3}A-3\cos A}}$

$\tt{\implies \dfrac{4\sin A-4\sin^{3} A}{4\cos^{3} A-2\cos A}}$

$\tt{\implies \dfrac{4\sin A(1-\sin^{2} A)}{2\cos A(2\cos^{2} A-1)}}$

$\tt{\implies \dfrac{4\sin A \;\cos^{2} A}{2\cos A\;\cos2A}}$

$\tt{\implies \dfrac{2\sin A\;\cos A}{\cos 2A}}$

$\tt{\implies \dfrac{\sin2A}{\cos2A}}$

$\tt{\implies \tan2A}$

Hence Proved