Question

1+cota+tana)(sina-cosa)=sec^3a-cosec^3a/sec^2a-cosec^2a

Answer 1

$\color {red}\huge\bold\star\underline\mathcal{Question:-}$ (1+cota+tana)(sina-cosa) = (sec^3a-cosec^3a)/(sec^2a*cosec^2a)

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:Answer}}}}}}}}}}$

Solving LHS ,

$(1+cota+tana)(sina-cosa) \\ \\ (1 + \frac{cosa}{sina} + \frac{sina}{cosa}) \: (sina-cosa) \\ \\ (sina + cosa + \frac{sin ^{2}a }{cosa} ) - (cosa + \frac{cos^{2}a }{sina} + sina) \\ \\ \frac{sin ^{2}a }{cosa} \: - \frac{cos^{2}a }{sina} \: \\ \\ \frac{sin^{2}a }{1} \times \frac{1}{cosa} - \frac{cos^{2}a }{1} \times \frac{1}{sina} \\ \\ \frac{seca}{cosec^{2}a } - \: \frac{coseca}{sec^{2}a } \\ \\ \frac{(sec^{3}a - cosec^{3}a)}{cosec^{2}a \times sec^{2} a }$

$\large\red{\boxed{\sf </strong><strong>LHS</strong><strong>=</strong><strong>RHS</strong><strong>}}$

$\huge\blue{</strong><strong>PROVED</strong><strong>}$