(D) Given the expression $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.Simplifying the terms inside the parentheses:$

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(D) Given the expression $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.Simplifying the terms inside the parentheses:$= \left(\frac{c+a+b}{c}\right) \left(\frac{b+c-a}{b}\right)$$= \frac{(b+c)+a}{c} \times \frac{(b+c)-a}{b}$$= \frac{(b+c)^2 - a^2}{bc}$$= \frac{b^2 + c^2 + 2bc - a^2}{bc}$$= \frac{b^2 + c^2 - a^2}{bc} + 2$Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,we have $\frac{b^2 + c^2 - a^2}{bc} = 2 \cos A$.$= 2 \cos A + 2$Given $\angle A = 60^{\circ}$,$\cos 60^{\circ} = 1/2$.$= 2(1/2) + 2 = 1 + 2 = 3$.

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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In a triangle $ABC$,with usual notations $\angle A=60^{\circ}$,then $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$

MHT CET 2022 All MHT CET

A
$3/2$
B
$1/2$
C
$1$
D
$3$

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Trigonometrical Equations

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Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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MHT CET 2022 Paper All MHT CET years →

Match the items of List-$I$ with those of List-$II$ (Here $\Delta$ denotes the area of $\triangle ABC$.)
List-$I$ List-$II$
$(A)$ $\sum \cot A$ $(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$ $(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$ $(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$ $(iv)$ $12 : 5 : 13$
$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
$(vi)$ $4\Delta$

Then the correct match is

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The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for

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In $\triangle ABC$,if $\sin^2 B = \sin C$ and $3 \cos^2 B = 2 \cos^2 C$,then $\triangle ABC$ is

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In triangle $ABC$,if $\frac{a+b+c}{BC+AB}+\frac{a+b+c}{AC+AB}=3$,then $\tan \frac{C}{8}=$

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The number of solutions of the equation $\left( 1 - \frac{1}{2 \sin x} \right) \cos^2 2x = 2 \sin x - 3 + \frac{1}{\sin x}$ in the interval $[0, 4\pi]$ is:

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List-$I$	List-$II$
$(A)$ $\sum \cot A$	$(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$	$(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$	$(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$	$(iv)$ $12 : 5 : 13$
	$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
	$(vi)$ $4\Delta$

In a triangle $ABC$,with usual notations $\angle A=60^{\circ}$,then $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$

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