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Question Number 84205 by 1406 last updated on 10/Mar/20
prove a. sin2x=tanx(1+cos2x) b. sin2x=((2tanx)/(1+tan^2 x))
$${prove}\: \\ $$$${a}.\:{sin}\mathrm{2}{x}={tanx}\left(\mathrm{1}+{cos}\mathrm{2}{x}\right) \\ $$$${b}.\:{sin}\mathrm{2}{x}=\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}} \\ $$
Answered by john santu last updated on 10/Mar/20
(b) sin 2x = 2sin xcos x ÷ ((cos^2 x)/(cos^2 x)) = ((2tan x)/(sec^2 x)) = ((2tan x)/(1+tan^2 x))
$$\left(\mathrm{b}\right)\:\mathrm{sin}\:\mathrm{2x}\:=\:\mathrm{2sin}\:\mathrm{xcos}\:\mathrm{x}\:\boldsymbol{\div}\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\frac{\mathrm{2tan}\:\mathrm{x}}{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}\:\:=\:\frac{\mathrm{2tan}\:\mathrm{x}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}} \\ $$
Answered by Kunal12588 last updated on 10/Mar/20
a. sin 2x = 2sin x cos x ⇒sin 2x = tan x(2 cos^2 x) ⇒sin 2x = tan x(1+cos2x) b. sin 2x = 2 sin x cos x ⇒sin 2x = ((2 sin x cos x)/(sin^2 x + cos^2 x)) ⇒sin 2x = (((2sin x cos x)/(cos^2 x))/((sin^2 x + cos^2 x)/(cos^2 x))) ⇒sin 2x = ((2 tan x)/(1+tan^2 x ))
$${a}.\:{sin}\:\mathrm{2}{x}\:=\:\mathrm{2}{sin}\:{x}\:{cos}\:{x} \\ $$$$\Rightarrow{sin}\:\mathrm{2}{x}\:=\:{tan}\:{x}\left(\mathrm{2}\:{cos}^{\mathrm{2}} \:{x}\right) \\ $$$$\Rightarrow{sin}\:\mathrm{2}{x}\:=\:{tan}\:{x}\left(\mathrm{1}+{cos}\mathrm{2}{x}\right) \\ $$$$ \\ $$$${b}.\:{sin}\:\mathrm{2}{x}\:=\:\mathrm{2}\:{sin}\:{x}\:{cos}\:{x}\: \\ $$$$\Rightarrow{sin}\:\mathrm{2}{x}\:=\:\frac{\mathrm{2}\:{sin}\:{x}\:{cos}\:{x}}{{sin}^{\mathrm{2}} \:{x}\:+\:{cos}^{\mathrm{2}} \:{x}}\: \\ $$$$\Rightarrow{sin}\:\mathrm{2}{x}\:=\:\frac{\frac{\mathrm{2}{sin}\:{x}\:{cos}\:{x}}{{cos}^{\mathrm{2}} \:{x}}}{\frac{{sin}^{\mathrm{2}} \:{x}\:+\:{cos}^{\mathrm{2}} \:{x}}{{cos}^{\mathrm{2}} \:{x}}}\: \\ $$$$\Rightarrow{sin}\:\mathrm{2}{x}\:=\:\frac{\mathrm{2}\:{tan}\:{x}}{\mathrm{1}+{tan}^{\mathrm{2}} \:{x}\:}\: \\ $$
Answered by $@ty@m123 last updated on 10/Mar/20
(a) RHS=tanx(1+cos2x) =tanx.2cos^2 x =((sin x)/(cos x)).2cos^2 x =2sin xcos x =sin 2x =LHS (b) RHS=((2tanx)/(1+tan^2 x)) =((2tan x)/(sec^2 x)) =((2((sin x)/(cos x)))/(1/(cos^2 x))) =2sin xcos x =sin 2x =LHS
$$\left({a}\right)\:{RHS}={tanx}\left(\mathrm{1}+{cos}\mathrm{2}{x}\right) \\ $$$$={tanx}.\mathrm{2cos}\:^{\mathrm{2}} {x} \\ $$$$=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}.\mathrm{2cos}\:^{\mathrm{2}} {x} \\ $$$$=\mathrm{2sin}\:{x}\mathrm{cos}\:{x} \\ $$$$=\mathrm{sin}\:\mathrm{2}{x} \\ $$$$={LHS} \\ $$$$\left({b}\right)\:{RHS}=\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{2tan}\:{x}}{\mathrm{sec}\:^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{2}\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}}{\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}} \\ $$$$=\mathrm{2sin}\:{x}\mathrm{cos}\:{x} \\ $$$$=\mathrm{sin}\:\mathrm{2}{x} \\ $$$$={LHS} \\ $$$$ \\ $$

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