MCQ
If $y = \sin (2{\sin ^{ - 1}}x),$ then ${{dy} \over {dx}} = $
- ✓${{2 - 4{x^2}} \over {\sqrt {1 - {x^2}} }}$
- B${{2 + 4{x^2}} \over {\sqrt {1 - {x^2}} }}$
- C${{2 - 4{x^2}} \over {\sqrt {1 + {x^2}} }}$
- D${{2 + 4{x^2}} \over {\sqrt {1 + {x^2}} }}$
==> $y = \sin 2\theta $
==>$\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{2\cos 2\theta }}{{\cos \theta }}$
$ = \frac{{2(1 - 2{{\sin }^2}\theta )}}{{\sqrt {1 - {{\sin }^2}\theta } }} = \frac{{2 - 4{x^2}}}{{\sqrt {1 - {x^2}} }}$.
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${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$ , ${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$
, ${{c}_{2}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{1}}}{|{{b}_{1}}{{|}^{2}}}{{b}_{1}}$
,${{c}_{3}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{2}}}{|{{b}_{2}}{{|}^{2}}}{{b}_{2}}$,
${{c}_{4}}=a-\frac{c.a}{|a{{|}^{2}}}a$.
Then which of the following is a set of mutually orthogonal vectors is