Question

Is LHS=RHS?

$\frac{{\tan }^2\theta }{1+{\tan }^2\theta }+\frac{{\cot }^3\theta }{1+{\cot }^2\theta }=\sec \theta sin\theta -2cosec\theta \cos \theta$

Say true or false.

• A. True
• B. False
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is B False

$\frac{co{t}^{3}x}{1+co{t}^{2}x}$
$=\frac{ta{n}^2\left(x\right).co{t}^{3}x}{1+ta{n}^2\left(x\right)}$
$=\frac{cot\left(x\right)}{1+ta{n}^2\left(x\right)}$
Hence the above expression becomes,
$\frac{ta{n}^{2}x+cotx}{1+ta{n}^2\left(x\right)}$
$=\frac{ta{n}^{3}x+1}{\left(tanx\right)(ta{n}^{2}x+1)}$
$=\frac{co{s}^3\left(x\right)(si{n}^{3}x+co{s}^{3}x)}{co{s}^{3}x.sinx}$
$=\frac{co{s}^{3}x+si{n}^{3}x}{sinx}$
RHS
$=\frac{1-2si{n}^{2}xco{s}^{2}x}{sinx.cosx}$
$=\frac{(si{n}^{2}x+co{s}^{2}x{)}^2-2si{n}^{2}xco{s}^{2}x}{sinx.cosx}$
$=\frac{si{n}^{4}x+co{s}^{4}x}{sinx.cosx}$
Hence $LHS\ne RHS$