Question

Question 5 (vii)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(vii) $\frac{(sin\theta -2si{n}^3\theta )}{(2co{s}^3\theta -cos\theta )}=tan\theta$

Answer 1

(vii) $\frac{(sin\theta -2si{n}^3\theta )}{(2co{s}^3\theta -cos\theta )}=tan\theta L.H.S.=\frac{(sin\theta -2si{n}^3\theta )}{(2co{s}^3\theta -cos\theta )}=\frac{\left[sin\theta \right(1-2si{n}^2\theta \left)\right]}{\left[cos\theta \right(2co{s}^2\theta -1\left)\right]}=\frac{sin\theta [1-2(1-co{s}^2\theta \left)\right]}{\left[cos\theta \right(2co{s}^2\theta -1\left)\right]}=\frac{\left[sin\theta \right(2co{s}^2\theta -1\left)\right]}{\left[cos\theta \right(2co{s}^2\theta -1\left)\right]}=\left[\frac{sin\theta }{cos\theta }\right]$
= $tan\theta =R.H.S.$