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Parin Sir > Quantitative Aptitude (Maths) > T-Ratios & Identities > Examples 1 to 5 of T-Ratios & Identities > Examples 1 to 5 of T-Ratios & Identities

Examples 1 to 5 of T-Ratios & Identities

August 11, 2020
Category: Examples 1 to 5 of T-Ratios & Identities

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(1) Convert $\frac{7\pi}{6}$ radians into degrees.

Soln. $\pi = 180$ ^o

$\left(\frac{7\pi}{6}\right)^c$ = $\left(\frac{180}{\pi}\times\frac{7\pi}{6}\right)^\circ$ = 210^o.

(2) Convert 75^o into radians.

Soln. 180^o = $\pi$ ^c

$\therefore$ 75^o = $\left(\frac{\pi}{180}\times75\right)^c$ = $\left(\frac{5\pi}{12}\right)$ radians.

(3) If sin $\theta = \frac{4}{5}$ and $\frac{\pi}{2} < \theta < \pi$ , find the values of cosθ, tanθ, cosecθ, secθ and cotθ.

Soln. Since $\frac{\pi}{2} < \theta < \pi$ , so θ lies in 2^nd quadrant.

In 2^nd quadrant, cosθ is negative.

⇒ secθ, tanθ, cotθ are also negative.

Now, sinθ = $\frac{4}{5}$ = $\frac{\mathrm{Opp.\ \ Side}}{\mathrm{Hypotenuse}} \Rightarrow$ AB = 4, AC = 5.

∴cos = $- \frac{\mathrm{Adj.\ \ Side}}{\mathrm{Hypotenuse}}$ = $- \frac{3}{5}.$

secθ = $\frac{1}{cos{\theta}} = - \frac{5}{3}$ and cosecθ = $\frac{1}{sin{\theta}} = \frac{5}{4}$ ;

tanθ = $\frac{sin{\theta}}{\cos{\theta}}$ = – $\frac{4}{3}$ and cotθ = $\frac{1}{tan{\theta}} = - \frac{3}{4}.$

Soln. Since $\frac{\pi}{2} < \theta < \pi$ , so θ lies in 2^nd quadrant.

In 2^nd quadrant, cosθ is negative.

$\therefore cos\theta = - \sqrt{1-{sin}^2{\theta}}$ = – $\sqrt{1-\frac{16}{25}}$ = – $\sqrt{\frac{9}{25}}$ = – $\frac{3}{5}.$

$sec\theta = \frac{1}{cos{\theta}}$ = – $\frac{5}{3}$ and $cosec\theta = \frac{1}{sin{\theta}}$ = $\frac{5}{4}$ ;

$tan\theta = \frac{sin{\theta}}{cos{\theta}}$ = – $\frac{4}{3}$ and $cot\theta = \frac{1}{tan{\theta}} = - \frac{3}{4}.$

(4) If cot θ = $\frac{5}{12}$ and $\pi < \theta < \frac{3\pi}{2}$ , find the values of cosec θ , tan θ, sec θ, cos θ and sin θ.

Soln. Since $\pi < \theta < \frac{3\pi}{2}$ , so lies in 3^rd quadrant.

In 3^rd quadrant, cosec θ , sec θ, cos θ sin θ are negative & tan θ is positive.

Now, $cot\theta = \frac{5}{12}$ = $\frac{\mathrm{Adj.\ \ Side}}{\mathrm{Opp.\ \ Side}}$ $\large \Rightarrow$ BC = 5, AB = 12.

$\therefore cos\theta = - \frac{\mathrm{Adj.\ \ Side}}{\mathrm{Hypotenuse}}$ = – $\frac{5}{13}$ & $\ sec\theta = \frac{1}{cos{\theta}}$ = – $\frac{13}{5}$ ;

$sin\theta = \frac{\mathrm{Opp.\ \ Side}}{\mathrm{Hypotenuse}}$ = – $\frac{12}{13}$ & $cosec\theta = \frac{1}{sin{\theta}}$ = – $\frac{13}{12}$ ;