Explanation
Simplifying the above equation we get
$$\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-cotA}$$
$$=\dfrac{\cos A}{1-\dfrac{\sin A}{\cos A}}+\dfrac{\sin A}{1-\dfrac{\cos A}{\sin A}}$$
$$=\dfrac{\cos ^2A}{\cos A-\sin A}+\dfrac{\sin ^2A}{\sin A-\cos A}$$
$$=\dfrac{\cos ^2A-\sin ^2A}{\cos A-\sin A}$$
$$=\dfrac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A-\sin A}$$
$$=\cos A+\sin A$$
Hence answer is B
Expanding the given expression we get
$$(\cos A+\sin A)\:^2-(\cos A+\sin A)\:^2$$
$$=\cos A\:^2+\sin A\:^2+2\cos A \sin A-(\cos A\:^2+\sin A\:^2-2\cos A \sin A)$$
$$=2(2\cos A \sin A)$$
$$=4 \cos A \sin A$$
Hence, option D is correct.
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)
$$=\dfrac{1}{\sin\alpha\:\tan\alpha\:\cos\alpha+\cos^2\alpha}$$
$$=\dfrac{1}{\sin^2\alpha+\cos^2\alpha}$$
$$=1$$
$$\cos A=2\sin A$$
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
$$cosec A=\pm\sqrt{5}$$
$$(\sin^2\theta+\cos^2\theta)\:^3=1$$
$$\sin^6\theta+\cos^6\theta+3\sin^2\theta \cos^2\theta(\sin^2\theta+\cos^2\theta)=1$$
$$\sin^6\theta+\cos^6\theta+3\sin^2\theta \cos^2\theta(1)=1$$
Hence answer is C
$$4 \cos ^2\theta -2\sqrt 2 \cos \theta-1=0$$ $$\cos \theta=\dfrac {2\sqrt 2\pm \sqrt {8+16}}{8}=\dfrac {\sqrt 2\pm \sqrt 6}{4}$$ $$\cos \theta =\dfrac {\sqrt 6+\sqrt 2}{4}$$ $$\Rightarrow \theta =\dfrac {\pi}{12}; 2\pi -\dfrac {\pi}{12}=\dfrac {23\pi}{12}$$ $$\cos \theta=-\dfrac {\sqrt 6-\sqrt 2}{4}$$ $$\cos \theta =\cos (\pi -\dfrac{5\pi}{12}); \cos (\pi +\dfrac{5\pi}{12})$$ $$\theta=\dfrac{7\pi}{12}; \dfrac{17\pi }{12}$$ Hence option B is the correct option
$$\alpha+\beta=90^0$$ ...(i)
$$\alpha-2\beta=0$$ ...(ii)
Subtracting ii from i we get
$$3\beta=90^0$$
$$\beta=30^0$$
$$\cos^2\alpha+\sin ^2\beta$$
$$=\cos^2(90^0-\beta)+\sin ^2\beta$$ ...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
$$n|\sin (x)|=m|\cos (x)|$$ $$\dfrac{n}{m}=|cot(x)|$$ Or $$\dfrac{m}{n}=|\tan (x)|$$ Or $$\tan (x)=\dfrac{m}{n}$$ Hence $$x=\tan ^{-1}(\dfrac{m}{n})$$ and $$x=\pi+\tan ^{-1}(\dfrac{m}{n})$$. And $$\tan (x)=\dfrac{-m}{n}$$ Hence $$x=-\tan ^{-1}(\dfrac{m}{n})$$ and $$x=\pi-\tan ^{-1}(\dfrac{m}{n})$$. Now if $$n,m\neq 0$$ we therefore get 4 solutions in the interval of $$[0,2\pi]$$ independent of the values of m and n.
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