Question Text
Question 1 :
If $\displaystyle \alpha +\beta =90^{\circ}$ and $ \alpha =2\beta $, then $ \cos ^{2}\alpha +\sin ^{2}\beta $ equals to
Question 2 :
The given expression is $\displaystyle \sin { \theta } \cos { \left( { 90 }^{ o }-\theta \right) } +\cos { \theta } \sin { \left( { 90 }^{ o }-\theta \right) } +4 $ equal to :<br/>
Question 3 :
If in $\triangle{ABC}$, $\angle {A} = 3\angle{B} $ then $\displaystyle sinB = \frac{1}{2} \sqrt\frac{3b-a}{b}$?
Question 4 :
$( \sec A + \tan A - 1 ) ( \sec A - \tan A + 1 ) =$
Question 5 :
$\cos { \left( \dfrac { 3\pi }{ 4 } +x \right) } -\cos { \left( \dfrac { 3\pi }{ 4 } -x \right) } =\sqrt { 2 } .\sin { x }$<br/>
Question 6 :
The value of $\sin { \left( {{ 45 }}^{o} +\theta \right) } -\cos { \left( {{ 45 }}^{o} -\theta \right) } $ is
Question 7 :
$\cos ^{ 2 }{ \cfrac { 3\pi }{ 5 } } +\cos ^{ 2 }{ \cfrac { 4\pi }{ 5 } } $ is equal to -
Question 9 :
In $\Delta ABC$ if $\dfrac{\cos A}{a} = \dfrac{\cos B}{b}$ then it is an isosceles triangle. (With the standard notations.)
Question 10 :
Assertion: $\sin 1^{\circ} < \cos 1^{\circ}$
Reason: $\sin \theta < \cos \theta$, when $0 < \theta < 90^{\circ}$