Math, asked by nandnishah11, 6 months ago

if tan (theta 1 + theta 2 ) =root 3 and sec (theta 1-theta 2) = 2/root 3 then find 2theta1 + tan 3theta2​

Answers

Answered by MaheswariS
10

\textbf{Given:}

\mathrm{tan(\theta_1+\theta_2)=\sqrt{3}}

\mathrm{sec(\theta_1-\theta_2)=\dfrac{2}{\sqrt{3}}}

\textbf{To find:}

\text{The value of}\,\,\mathrm{2\theta_1+3\theta_2}

\textbf{Solution:}

\text{Consider,}

\mathrm{tan(\theta_1+\theta_2)=\sqrt{3}}

\implies\mathrm{\theta_1+\theta_2=60^\circ}- - - - - -(1)

\mathrm{sec(\theta_1-\theta_2)=\dfrac{2}{\sqrt{3}}}

\text{Taking reciprocals}

\mathrm{cos(\theta_1-\theta_2)=\dfrac{\sqrt{3}}{2}}

\implies\mathrm{\theta_1-\theta_2=30^\circ}- - - - - - (2)

\mathrm{\theta_1+\theta_2=60^\circ}- - - - - -(1)

\mathrm{\theta_1-\theta_2=30^\circ}- - - - - - (2)

\mathrm{Adding\;(1)\;and\;(2),\;we\;get}

\mathrm{2\theta_1=90^\circ}

\mathrm{\theta_1=\dfrac{90^\circ}{2}}

\implies\boxed{\mathrm{\theta_1=45^\circ}}

\mathrm{(1)\implies}

\mathrm{45^\circ+\theta_2=60^\circ}

\mathrm{\theta_2=60^\circ-45^\circ}

\implies\boxed{\mathrm{\theta_2=15^\circ}}

\text{Now,}

\mathrm{2\,\theta_1+3\,\theta_2}

\mathrm{=2{\times}30^\circ+3{\times}15^\circ}

\mathrm{=60^\circ+45^\circ}

\mathrm{=105^\circ}

\implies\boxed{\mathrm{2\,\theta_1+3\,\theta_2=105^\circ}}

Answered by anshukumarprajapati4
0

an(θ

1

2

)=

3

\implies\mathrm{\theta_1+\theta_2=60^\circ}⟹θ

1

2

=60

- - - - - -(1)

\mathrm{sec(\theta_1-\theta_2)=\dfrac{2}{\sqrt{3}}}sec(θ

1

−θ

2

)=

3

2

\text{Taking reciprocals}Taking reciprocals

\mathrm{cos(\theta_1-\theta_2)=\dfrac{\sqrt{3}}{2}}cos(θ

1

−θ

2

)=

2

3

\implies\mathrm{\theta_1-\theta_2=30^\circ}⟹θ

1

−θ

2

=30

- - - - - - (2)

\mathrm{\theta_1+\theta_2=60^\circ}θ

1

2

=60

- - - - - -(1)

\mathrm{\theta_1-\theta_2=30^\circ}θ

1

−θ

2

=30

- - - - - - (2)

\mathrm{Adding\;(1)\;and\;(2),\;we\;get}Adding(1)and(2),weget

\mathrm{2\theta_1=90^\circ}2θ

1

=90

\mathrm{\theta_1=\dfrac{90^\circ}{2}}θ

1

=

2

90

\implies\boxed{\mathrm{\theta_1=45^\circ}}⟹

θ

1

=45

\mathrm{(1)\implies}(1)⟹

\mathrm{45^\circ+\theta_2=60^\circ}45

2

=60

\mathrm{\theta_2=60^\circ-45^\circ}θ

2

=60

−45

\implies\boxed{\mathrm{\theta_2=15^\circ}}⟹

θ

2

=15

\text{Now,}Now,

\mathrm{2\,\theta_1+3\,\theta_2}2θ

1

+3θ

2

\mathrm{=2{\times}30^\circ+3{\times}15^\circ}=2×30

+3×15

\mathrm{=60^\circ+45^\circ}=60

+45

\mathrm{=105^\circ}=105

\implies\boxed{\mathrm{2\,\theta_1+3\,\theta_2=105^\circ}}⟹

1

+3θ

2

=105

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