Question

if tan =√3. find the all the T.ratio
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Answer 1

Answer for students for class 10.

All trigonometric ratios for θ if $\sf{tan\:\theta=\sqrt{3} \:(2n\pi \leq \theta\leq \dfrac{\pi }{2} +2n\pi )}$

• $\sf{sin\:\theta=\dfrac{\sqrt{3} }{2} }$, $\sf{cos\:\theta=\dfrac{1}{2} }$, $\sf{tan\:\theta=\sqrt{3} }$ (Basic Ratios)
• $\sf{csc\:\theta=\dfrac{2}{\sqrt{3} } }$, $\sf{sec\:\theta=2}$, $\sf{cot\:\theta=\dfrac{1}{3} }$ (Inverse Ratios)

 

Answer for students over class 10.

All trigonometric ratios for θ if $\sf{tan\:\theta=\sqrt{3} \:(\pi +2n\pi\leq \theta\leq \dfrac{3}{4} \pi+2n\pi)}$

• $\sf{sin\:\theta=-\dfrac{\sqrt{3} }{2} }$, $\sf{cos\:\theta=-\dfrac{1}{2} }$, $\sf{tan\:\theta=\sqrt{3} }$ (Basic Ratios)
• $\sf{csc\:\theta=-\dfrac{2}{\sqrt{3} } }$, $\sf{sec\:\theta=-2}$, $\sf{cot\:\theta=\dfrac{1}{3} }$ (Inverse Ratios)

 

Let's find the values using trigonometric identities.

 

We know that $\sf{sin^2\:\theta+cos^2\:\theta=1}$.

Let's divide both sides by $\sf{cos^2\:\theta}$.

$\implies\sf{\dfrac{sin^2\:\theta}{cos^2\:\theta}+\dfrac{cos^2\:\theta}{cos^2\:\theta}=\dfrac{1}{cos^2\:\theta} }$

$\implies\sf{tan^2\:\theta+1=\dfrac{1}{cos^2\:\theta} }$

 

Given $\sf{tan\:\theta=\sqrt{3} }$

$\implies\sf{(\sqrt{3} )^2+1=\dfrac{1}{cos^2\:\theta} }$

$\implies\sf{4=\dfrac{1}{cos^2\:\theta}}$

$\implies\sf{cos^2\:\theta=\dfrac{1}{4} \:(0\leq cos\:\theta\leq 1)}$

Therefore cosine is $\dfrac{1}{2}$.

 

Let's find sine value using the identity.

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

$\implies\sf{sin^2\:\theta+(\dfrac{1}{2} )^2=1}$

$\implies\sf{sin^2\:\theta=\dfrac{3}{4} \:(0\leq sin\:\theta\leq 1)}$

Therefore sine is $\dfrac{\sqrt{3} }{2}$.

 

The cosecant, secant, and cotangent values are multiplicative inverses of sin θ, cos θ, and tan θ.

 

More information:

Common mistakes?

We take $\sf{sin^2\:\theta=(sin\:\theta)^2}$, but don't take $\sf{sin^{-1}\:\theta=\dfrac{1}{sin\:\theta} }$. The inverse function is not a multiplicative inverse.

 

Trigonometric identity?

$\sf{a^2+b^2=c^2}$ (Pythagoras Theorem)

$\implies\sf{\dfrac{a^2}{c^2} +\dfrac{b^2}{c^2} =\dfrac{c^2}{c^2} }$

$\implies\sf{(\dfrac{a}{c} )^2+(\dfrac{b}{c} )^2=1}$

The two values are either sine or cosine. From this, we get a trigonometric identity below.

$\therefore\sf{sin^2\:\theta+cos^2\:\theta=1}$