If tan-3- then sec - -a 23b 32c 12d 2

(d) 2 Let us first draw a right ∆ABC right angled at B and ∠A=θ. Given: tan θ = 3 But tan θ = BCAB So, BCAB = 31 Thus, BC = 3k and AB = k Using Pythagoras th ...

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Standard X

Mathematics

Trigonometric Ratios

If tanθ=3, th...

Question

If tan θ = $\sqrt{3}$ , then sec θ = ?
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 2

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Solution

(d) 2

Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: tan θ = $\sqrt{3}$
But tan θ = $\frac{B C}{A B}$
So, $\frac{B C}{A B}$ = $\frac{\sqrt{3}}{1}$
Thus, BC = $\sqrt{3}$ k and AB = k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = ( $\sqrt{3}$ k)² + (k)²
⇒ AC²= 4k²
⇒ AC = 2k
∴ sec θ = $\frac{AC}{AB} = \frac{2 k}{k} = \frac{2}{1}$

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If tan θ = 3, then sec θ = ? (a) 23 (b) 32 (c) 12 (d) 2

(d) 2 Let us first draw a right ∆ABC right angled at B and ∠A=θ. Given: tan θ = 3 But tan θ = BCAB So, BCAB = 31 Thus, BC = 3k and AB = k Using Pythagoras theorem, we get: AC2 = AB2 + BC2 ⇒ AC2 = (3 k)2 + (k)2 ⇒ AC2= 4k2 ⇒ AC = 2k ∴ sec θ = ACAB = 2kk = 21

If tan θ = $\sqrt{3}$ , then sec θ = ?
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 2