Question:
$ \tan\left(\frac{\pi}{4} +\frac{\theta}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right)$ is equal to
$ \tan\left(\frac{\pi}{4} +\frac{\theta}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right)$ is equal to
Updated On: Jun 7, 2024
- $\sec \, \theta $
- $2 \sec \, \theta $
- $\sec \, \frac {\theta} {2} $
- $\sin \, \theta $
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The Correct Option is B
Solution and Explanation
We have,
$ \tan \left(\frac{\pi}{4}+\right. \left.\frac{\theta}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right) $
$= \frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}+\frac{\tan \frac{\pi}{4}-\tan \frac{\theta}{2}}{1+\tan \frac{\pi}{4} \tan \frac{\theta}{2}} $
$= \frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}+\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}} $
$= \frac{\left(1+\tan \frac{\theta}{2}\right)^{2}+\left(1-\tan \frac{\theta}{2}\right)^{2}}{1-\tan ^{2} \frac{\theta}{2}} $
$=\frac{1+\tan ^{2} \frac{\theta}{2}+2 \tan \frac{\theta}{2}+1+\tan ^{2} \frac{\theta}{2}-2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}$
$=2\left[\frac{1+\tan ^{2} \theta / 2}{1-\tan ^{2} \theta / 2}\right]$
$=\frac{2}{\cos \theta} \,\,\,\left[\because \cos 2 \,\theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right]$
$=2 \sec \,\theta$
$ \tan \left(\frac{\pi}{4}+\right. \left.\frac{\theta}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right) $
$= \frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}+\frac{\tan \frac{\pi}{4}-\tan \frac{\theta}{2}}{1+\tan \frac{\pi}{4} \tan \frac{\theta}{2}} $
$= \frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}+\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}} $
$= \frac{\left(1+\tan \frac{\theta}{2}\right)^{2}+\left(1-\tan \frac{\theta}{2}\right)^{2}}{1-\tan ^{2} \frac{\theta}{2}} $
$=\frac{1+\tan ^{2} \frac{\theta}{2}+2 \tan \frac{\theta}{2}+1+\tan ^{2} \frac{\theta}{2}-2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}}$
$=2\left[\frac{1+\tan ^{2} \theta / 2}{1-\tan ^{2} \theta / 2}\right]$
$=\frac{2}{\cos \theta} \,\,\,\left[\because \cos 2 \,\theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right]$
$=2 \sec \,\theta$
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Concepts Used:
Properties of Inverse Trigonometric Functions
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Property Set 1:
- Sin−1(x) = cosec−1(1/x), x∈ [−1,1]−{0}
- Cos−1(x) = sec−1(1/x), x ∈ [−1,1]−{0}
- Tan−1(x) = cot−1(1/x), if x > 0 (or) cot−1(1/x) −π, if x < 0
- Cot−1(x) = tan−1(1/x), if x > 0 (or) tan−1(1/x) + π, if x < 0
Property Set 2:
- Sin−1(−x) = −Sin−1(x)
- Tan−1(−x) = −Tan−1(x)
- Cos−1(−x) = π − Cos−1(x)
- Cosec−1(−x) = − Cosec−1(x)
- Sec−1(−x) = π − Sec−1(x)
- Cot−1(−x) = π − Cot−1(x)
Property Set 3:
- Sin−1(1/x) = cosec−1x, x≥1 or x≤−1
- Cos−1(1/x) = sec−1x, x≥1 or x≤−1
- Tan−1(1/x) = −π + cot−1(x)
Property Set 4:
- Sin−1(cos θ) = π/2 − θ, if θ∈[0,π]
- Cos−1(sin θ) = π/2 − θ, if θ∈[−π/2, π/2]
- Tan−1(cot θ) = π/2 − θ, θ∈[0,π]
- Cot−1(tan θ) = π/2 − θ, θ∈[−π/2, π/2]
- Sec−1(cosec θ) = π/2 − θ, θ∈[−π/2, 0]∪[0, π/2]
- Cosec−1(sec θ) = π/2 − θ, θ∈[0,π]−{π/2}
- Sin−1(x) = cos−1[√(1−x2)], 0≤x≤1 = −cos−1[√(1−x2)], −1≤x<0
Property Set 5:
- Sin−1x + Cos−1x = π/2
- Tan−1x + Cot−1(x) = π/2
- Sec−1x + Cosec−1x = π/2
Property Set 6:
- If x, y > 0
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
- If x, y < 0
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
Property Set 7:
- sin−1(x) + sin−1(y) = sin−1[x√(1−y2)+ y√(1−x2)]
- cos−1x + cos−1y = cos−1[xy−√(1−x2)√(1−y2)]
Property Set 8:
- sin−1(sin x) = −π−π, if x∈[−3π/2, −π/2]
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
- cos−1(cos x) = 2π+x, if x∈[−2π,−π]
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
- tan−1(tan x) = π+x, x∈(−3π/2, −π/2)
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)
Property Set 9:
