MCQ
$\sin \theta + \cos \theta $ ની કિમત મહતમ થવા માટે . . . .
- A$\theta = {30^o}$
- ✓$\theta = {45^o}$
- C$\theta = {60^o}$
- D$\theta = {90^o}$
But $ - 1 \le \sin \left( {\theta + \frac{\pi }{2}} \right) \le 1$
$\Rightarrow - \sqrt 2 \le \sqrt 2 \sin \left( {\theta + \frac{\pi }{4}} \right) \le \sqrt 2 $.
Hence the maximum value of $(\sin \theta + \cos \theta )$
$i.e.$, of $\sqrt 2 \sin \left( {\theta + \frac{\pi }{4}} \right) = \sqrt 2 $.
$\therefore $$\sin \left( {\theta + \frac{\pi }{4}} \right) = 1 $
$\Rightarrow \sin \left( {\theta + \frac{\pi }{4}} \right) = \sin \frac{\pi }{2}$
==> $\theta + \frac{\pi }{4} = \frac{\pi }{2} $
$\Rightarrow \theta = \frac{\pi }{4} = {45^o}$.
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કારણ ${\rm{(R)}}$ બિંદુઓ ${\rm{ (}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{)}}$ એઅતિવલય ${\rm{ }}\,\,\frac{{{x^2}}}{{{a^2}}}\, - \,\,\frac{{{y^2}}}{{{b^2}}}\, = \,\,1$ ની અંદર આવેલું , તો $\frac{{x_{^1}^2}}{{{a^2}}}\, - \,\,\frac{{y_1^2}}{{{b^2}}}\, - \,\,1\,\, < \,\,0$