The-angles-A-B-C-of-a-triangle-ABC-satisfy-4cosAcosB-sin2A-sin2B-sin2C-4-Then-which-of-the-following-statements-is-are-correct-1-The-triangle-ABC-is-right-angled-2-The-triangle-ABC-is-

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Question Number 18063 by Tinkutara last updated on 14/Jul/17

$$\mathrm{The}\mathrm{angles}A,B,C\mathrm{of}\mathrm{a}\mathrm{triangle}ABC$$$$\mathrm{satisfy}4\cos A\cos B+\sin 2A+\sin 2B+$$$$\sin 2C=4.\mathrm{Then}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}$$$$\mathrm{statements}\mathrm{is}/\mathrm{are}\mathrm{correct}?$$$$\left(1\right)\mathrm{The}\mathrm{triangle}ABC\mathrm{is}\mathrm{right}\mathrm{angled}$$$$\left(2\right)\mathrm{The}\mathrm{triangle}ABC\mathrm{is}\mathrm{isosceles}$$$$\left(3\right)\mathrm{The}\mathrm{triangle}ABC\mathrm{is}\mathrm{neither}$$$$\mathrm{isosceles}\mathrm{nor}\mathrm{right}\mathrm{angled}$$$$\left(4\right)\mathrm{The}\mathrm{triangle}ABC\mathrm{is}\mathrm{equilateral}$$

Commented by prakash jain last updated on 14/Jul/17

$$\sin 2A+\sin 2B+\sin 2C$$$$=2\sin (A+B)\cos (A-B)+2\sin C\cos C$$$$=2\sin C\cos (A-B)-2\sin C\cos (A+B)$$$$4\cos A\cos B=2\cos (A+B)+2\cos (A-B)$$$$givenexpression.$$$$-\cos (A+B)(1-\sin C)$$$$+\cos (A-B)(1+\sin C)=2$$$$case1:if(A+B)<\frac{\pi }{2},\sin C<1$$$$\cos (A-B)(1+\sin C)<2$$$$\mathrm{LHS}=2-(some+venumber)\ne 2$$$$case2:if(A+B)⩾\pi /2$$$$A+B=\frac{\pi }{2}+x\Rightarrow C=\frac{\pi }{2}-x$$$$\sin x(1-\cos x)+\cos (A-B)(1+\cos x)=2$$$$\cos (A-B)=\frac{2-\sin x+\sin x\cos x}{1+\cos x}⩾1$$$$\mathrm{RHS}=1iffx=0$$$$\cos (A-B)=1\Rightarrow A=B$$$$\Rightarrow A+B=\frac{\pi }{2}\Rightarrow C=\frac{\pi }{2}$$$$so\mathrm{triangle}\mathrm{is}\mathrm{right}\mathrm{angles}\mathrm{and}$$$$\mathrm{isoceles}.$$

Commented by prakash jain last updated on 15/Jul/17

$$\frac{2-\sin x+\sin x\cos x}{1+\cos x}$$$$=\frac{2-\sin x(1-\cos x)}{1+\cos x}$$$$=\frac{{2\cos }^2\frac{x}{2}+{2\sin }^2\frac{x}{2}-2\sin x{\sin }^2\frac{x}{2}}{{2\cos }^2\frac{x}{2}}$$$$=1+\frac{{\sin }^2\frac{x}{2}(1-\sin x)}{{\cos }^2\frac{x}{2}}⩾1$$

Commented by Tinkutara last updated on 15/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Answered by Tinkutara last updated on 16/Jul/17

$$\mathrm{In}\mathrm{any}\mathrm{\Delta }ABC,\sin 2A+\sin 2B+$$$$\sin 2C=4\sin A\sin B\sin C$$$$\therefore 4\cos A\cos B+4\sin A\sin B\sin C=4$$$$\sin C=\frac{1-\cos A\cos B}{\sin A\sin B}⩽1\dots \left(\mathit{i}\right)$$$$1⩽\cos A\cos B+\sin A\sin B=\cos (A-B)$$$$\therefore A=B\mathrm{and}\mathrm{putting}\mathrm{this}\mathrm{in}\left(\mathit{i}\right),$$$$\sin C=1\Rightarrow C=90°.$$$$\mathrm{Hence}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{isosceles}$$$$\mathrm{triangle}.$$

Commented by prakash jain last updated on 16/Jul/17

$$\mathrm{Simple}\mathrm{approach}.$$

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