Trigonometric Functions Test -1

$$\alpha+\beta=90^0$$ ...(i)

$$\alpha-2\beta=0$$ ...(ii)

Subtracting ii from i we get

$$3\beta=90^0$$

$$\beta=30^0$$

Therefore

$$\cos^2\alpha+\sin ^2\beta$$

$$=\cos^2(90^0-\beta)+\sin ^2\beta$$ ...from(i)

$$=\sin ^2\beta+\sin ^2\beta$$

$$=2\sin ^2\beta$$

Substituting the value of $$\beta$$ we get

$$2\sin ^2\beta=2(\dfrac{1}{2})\:^2$$

$$=\dfrac{1}{2}$$

Hence the answer is A

As we know $$AM >= GM$$
$$\sin x=x+\dfrac {1}{x}, x\epsilon (0, 2\pi)$$R. H. S, $$x+\dfrac {1}{x}\geq 2$$
While $$\sin x\leq 1$$
Option B

$$\sin\beta=\dfrac{12}{13}$$

By expanding the expression we get

$$(\sin\theta+\csc\theta)\:^2+(\cos\theta+sec\theta)\:^2-(\tan^2\theta+cot^2\theta)$$

$$=\sin^2\theta+\csc^2\theta+2+sec^2\theta+\cos^2\theta+2-\tan^2\theta-cot^2\theta$$

$$=4+(\sin^2\theta+\cos^2\theta)+(\csc^2\theta-cot^2\theta)+(sec^2\theta-\tan^2\theta)$$

$$=4+1+1+1=7$$

Hence answer is D

Simplifying the given expression we get

$$\dfrac{\dfrac{\sin\alpha}{\cos\alpha}}{\sin^2\alpha\:\tan\alpha+\sin\alpha \cos\alpha}$$

$$=\dfrac{\sin\alpha}{(\sin\alpha\:\cos\alpha)(\sin\alpha\:\tan\alpha+\cos\alpha)}$$ ...(taking $$\sin\alpha$$ common from the denominator)

$$=\dfrac{1}{\sin\alpha\:\tan\alpha\:\cos\alpha+\cos^2\alpha}$$

$$=\dfrac{1}{\sin^2\alpha+\cos^2\alpha}$$

$$=1$$

Hence answer is D

$$\cos A=2\sin A$$

Dividing by $$\sin A $$on both the sides we get

$$\cot A=2$$ ...(i)

$$cosec^2A=1+\cot^2A$$

$$cosec^2A=1+2^2$$ ...(from i )

$$cosec^2A=5$$

Therefore

$$cosec A=\pm\sqrt{5}$$

Hence answer is B

$$(\sin^2\theta+\cos^2\theta)\:^3=1$$

$$\sin^6\theta+\cos^6\theta+3\sin^2\theta \cos^2\theta(\sin^2\theta+\cos^2\theta)=1$$

$$\sin^6\theta+\cos^6\theta+3\sin^2\theta \cos^2\theta(1)=1$$

Hence answer is C

$$\sin x. \cos y =1 $$
$$ \Rightarrow \sin x=1 , \cos y =1 $$ or $$\sin x =- 1 , \cos y = -1$$
Hence, the ordered pairs of $$(x,y)$$ are $$(\displaystyle \frac {\pi}{2} , 0)$$ ,  $$ (\displaystyle \frac {\pi}{2} , 2\pi)$$ ,$$ ( \displaystyle \frac {3\pi}{2} , \pi) $$

$$4 \cos ^2\theta -2\sqrt 2 \cos \theta-1=0$$
$$\cos \theta=\dfrac  {2\sqrt 2\pm \sqrt {8+16}}{8}=\dfrac  {\sqrt 2\pm \sqrt 6}{4}$$
$$\cos \theta =\dfrac  {\sqrt 6+\sqrt 2}{4}$$
$$\Rightarrow \theta =\dfrac  {\pi}{12}; 2\pi -\dfrac  {\pi}{12}=\dfrac  {23\pi}{12}$$
$$\cos \theta=-\dfrac  {\sqrt 6-\sqrt 2}{4}$$
$$\cos \theta =\cos (\pi -\dfrac{5\pi}{12}); \cos (\pi +\dfrac{5\pi}{12})$$
$$\theta=\dfrac{7\pi}{12}; \dfrac{17\pi }{12}$$
Hence option B is the correct option