Question

(i) ${\sin }^6\mathrm{\theta }+{\cos }^6\mathrm{\theta }=1-3{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }$
(ii) ${\sin }^2\mathrm{\theta }+{\cos }^4\mathrm{\theta }={\cos }^2\mathrm{\theta }+{\sin }^4\mathrm{\theta }$
(iii) ${\mathrm{cosec}}^4\mathrm{\theta }-{\mathrm{cosec}}^2\mathrm{\theta }={\cot }^4\mathrm{\theta }+{\cot }^2\mathrm{\theta }$

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \left(\text{i}\right){\text{LHS=sin}}^6\theta +{\text{cos}}^6\theta \\ \text{=}{\left({\text{sin}}^2\theta \right)}^3+{\left({\text{cos}}^2\theta \right)}^3\\ \text{=}({\text{sin}}^2\theta +{\text{cos}}^2\theta )({\text{sin}}^4\theta -{\text{sin}}^2\theta {\text{cos}}^2\theta +{\text{cos}}^4\theta )\\ \text{=1×{}{\left({\text{sin}}^2\theta \right)}^2+2{\text{sin}}^2\theta {\text{cos}}^2\theta +{\left({\text{cos}}^2\theta \right)}^2-3{\text{sin}}^2\theta {\text{cos}}^2\theta \}\\ \text{=}{({\text{sin}}^2\theta +{\text{cos}}^2\theta )}^2-3{\text{sin}}^2\theta {\text{cos}}^2\theta \\ \text{=}{\left(1\right)}^2-3{\text{sin}}^2\theta {\text{cos}}^2\theta \\ \text{=1}-3{\text{sin}}^2\theta {\text{cos}}^2\theta \\ \text{=RHS}\end{array} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \\ \begin{array}{l}\left(\text{ii}\right){\text{LHS=sin}}^2\mathrm{\theta }+{\text{cos}}^4\mathrm{\theta }\\ {\text{=sin}}^2\mathrm{\theta }+{\left({\text{cos}}^2\mathrm{\theta }\right)}^2\\ {\text{=sin}}^2\mathrm{\theta }+{(1-{\text{sin}}^2\mathrm{\theta })}^2\\ {\text{=sin}}^2\mathrm{\theta }+1-2{\text{sin}}^2\mathrm{\theta }+{\text{sin}}^4\mathrm{\theta }\\ \text{=1}-{\text{sin}}^2\mathrm{\theta }+{\text{sin}}^4\mathrm{\theta }\\ {\text{=cos}}^2\mathrm{\theta }+{\text{sin}}^4\mathrm{\theta }\\ \text{=RHS}\end{array} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \\ \begin{array}{l}\left(\text{iii}\right){\text{LHS=cosec}}^4\mathrm{\theta }-{\text{cosec}}^2\mathrm{\theta }\\ {\text{=cosec}}^2\mathrm{\theta }({\text{cosec}}^2\mathrm{\theta }-1)\\ {\text{=cosec}}^2\mathrm{\theta }\times {\text{cot}}^2\mathrm{\theta }(\because {\text{cosec}}^2\mathrm{\theta }-{\text{cot}}^2\mathrm{\theta }=1)\\ \text{=}(1+{\text{cot}}^2\mathrm{\theta }){\text{×cot}}^2\mathrm{\theta }\\ {\text{=cot}}^2\mathrm{\theta }+{\text{cot}}^4\mathrm{\theta }\\ \text{=RHS}\end{array} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$