if tan (theta 1 + theta 2 ) =root 3 and sec (theta 1-theta 2) = 2/root 3 then find 2theta1 + tan 3theta2
Answers
- - - - - -(1)
- - - - - - (2)
- - - - - -(1)
- - - - - - (2)
an(θ
1
+θ
2
)=
3
\implies\mathrm{\theta_1+\theta_2=60^\circ}⟹θ
1
+θ
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
+θ
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
∘
+θ
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}}⟹
2θ
1
+3θ
2
=105
∘