Question

Using trigonometric identities $\frac{{\cot }^3\theta }{1+{\cot }^2\theta }+\frac{{\tan }^3\theta }{1+{\tan }^2\theta }=\sec \theta .\csc \theta$

• A. True
• B. False

Answer 1

The correct option is B False

Taking LHS

$\frac{co{t}^3\theta }{1+co{t}^2\theta }+\frac{ta{n}^3\theta }{1+ta{n}^2\theta }$

$=\frac{co{t}^3\theta }{cose{c}^2\theta }+\frac{ta{n}^3\theta }{1+ta{n}^2\theta }$

$=\frac{co{s}^3\theta }{sin\theta }+\frac{si{n}^3\theta }{cos\theta }$

$=\frac{si{n}^4\theta +co{s}^4\theta }{sin\theta .cos\theta }$

$=\frac{si{n}^4\theta +co{s}^4\theta +2si{n}^2\theta co{s}^2\theta -2co{s}^2\theta si{n}^2\theta }{sin\theta .cos\theta }$

$=\frac{(si{n}^2\theta +co{s}^2\theta {)}^2-2si{n}^2\theta co{s}^2\theta }{sin\theta .cos\theta }$

$=sec\theta cosec\theta -2sin\theta cos\theta$

$\ne RHS$