Question

prove that:
plzz help me out​

Answer 1

$\underline{\underline{\textsf{\textbf{\purple{ \mapsto PROOF:}}}}}$

$\underline{\red{\sf \dashrightarrow LHS }}\:=\:\tan\bigg(\dfrac{\pi}{4}+\theta\bigg)$

$\underline{\red{\sf \dashrightarrow RHS }}\:=\:\tan\bigg(\dfrac{1-\tan\theta}{1+\tan\theta}\bigg)$

Simplifying LHS :-

= tan ( π/4 + θ )

= tan ( 180°/4 + θ )

= tan ( 45° + θ )

= tan45° + tanθ / 1-tan45°. tanθ

$\boxed{\green{\bf \dag tan(A+B)=\dfrac{tanA+tanB}{1-tanA.tanB}}}$

= 1 + tanθ / 1 - 1 . tanθ

= 1 + tanθ / 1 - tanθ.

= RHS .

Hence Proved :)

$\rule{200}3$

$\underline{\underline{\textsf{\textbf{\purple{ \mapsto MorE\:To\;KnoW:}}}}}$

Some more related formula ,

➳ $\sf\red{ cosec^2\theta - cot^2\theta=1}$

➳ $\sf\red{ sin^2\theta + cos^2\theta=1}$

➳ $\sf\red{ sec^2\theta - tan^2\theta=1}$

➳ $\sf\red{ sin(A+B)=sinA.cosB+cosA.sinB}$

➳ $\sf\red{ sin(A-B)=sinA.cosB-cosA.sinB}$

➳ $\sf\red{cos(A+B)=cosA.cosB-sinA.sinB }$

➳ $\sf\red{cos(A-B)=cosA.cosB+sinA.sinB }$

➳ $\sf\red{ tan(A+B)=\dfrac{tanA+tanB}{1-tanA.tanB}}$

➳ $\sf\red{ tan(A-B)=\dfrac{tanA-tanB}{1+tanA.tanB}}$

➳ $\sf\red{ tan(A+B+C)=\dfrac{tanA+tanB+tanC-tanA.tanB.tanC}{1-tanA.tanB-tanB.tanC-tanC.tanA}}$

$\rule{200}3$

Values of trignometric Ratios for some special angles .

$\begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}$