Reciprocal identities express the inverse relationships between the six trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent.
Formulas
Here are the three basic trigonometric functions, sine, cosine, and tangent, and their relation to their reciprocals:
Sine and Cosec
Sine is the reciprocal of cosec and vice versa.
- ${\sin \theta =\dfrac{1}{\text{cosec}\, \theta }}$
- ${\text{cosec}\, \theta =\dfrac{1}{\sin \theta }}$
Cosine and Sec
Cosine is the reciprocal of sec and vice versa.
- ${\cos \theta =\dfrac{1}{\sec \theta }}$
- ${\sec \theta =\dfrac{1}{\cos \theta }}$
Tangent and Cotangent
Tangent is the reciprocal of cotangent and vice versa.
- ${\tan \theta =\dfrac{1}{\cot \theta }}$
- ${\cot \theta =\dfrac{1}{\tan \theta }}$
Proofs
Here, we will consider a right-angled triangle ABC, which is right-angled at C, to derive the reciprocals.
Sine and Cosec
From the triangle ABC,
sin θ = ${\dfrac{b}{c}}$ = ${\dfrac{Opposite \ side}{Hypotenuse}}$
As we know,
The reciprocal of ${\dfrac{b}{c}}$ is ${\dfrac{c}{b}}$
⇒ ${\dfrac{c}{b}}$ = ${\dfrac{Hypotenuse}{Opposite \ side}}$ = cosec θ
Thus, sin θ is the reciprocal of cosec θ and vice-versa.
Cosine and Sec
Similarly, from the triangle ABC,
cos θ = ${\dfrac{a}{c}}$ = ${\dfrac{Adjacent \ side}{Hypotenuse}}$
As we know,
The reciprocal of ${\dfrac{a}{c}}$ is ${\dfrac{c}{a}}$
⇒ ${\dfrac{c}{a}}$ = ${\dfrac{Hypotenuse}{Adjacent \ side}}$ = sec θ
Thus, cos θ is the reciprocal of sec θ and vice-versa.
Tangent and Cotangent
Again, from the triangle ABC,
tan θ = ${\dfrac{b}{a}}$ = ${\dfrac{Opposite \ side}{Adjacent \ side}}$
As we know,
The reciprocal of ${\dfrac{b}{a}}$ is ${\dfrac{a}{b}}$
⇒ ${\dfrac{a}{b}}$ = ${\dfrac{Adjacent \ side}{Opposite \ side}}$ = cot θ
Thus, tan θ is the reciprocal of cot θ and vice-versa.
Relationship Between Reciprocal Identities
The product of a trigonometric function and its reciprocal is always equal to 1:
- sin θ × cosec θ = 1
- cos θ × sec θ = 1
- tan θ × cot θ = 1
Finding Values Using Reciprocal Identities
Find: cosec θ if sin θ = ${\dfrac{3}{5}}$
As we know, cosec θ is the reciprocal of sin θ
Here, ${\text{cosec}\, \theta =\dfrac{1}{\sin \theta }}$ and sin θ = ${\dfrac{3}{5}}$
Thus, cosec θ = ${\dfrac{5}{3}}$
Find the value of cot 60° if tan 60° = ${\sqrt{3}}$
As we know, cot θ is the reciprocal of tan θ
Here, cot 60° = ${\dfrac{1}{\tan 60^{\circ }}}$ and tan 60° = ${\sqrt{3}}$
Thus, cot 60° = ${\dfrac{1}{\sqrt{3}}}$
Simplifying Using Reciprocal Identities
The reciprocal trigonometric functions can also be used to simplify complex mathematical expressions.
Simplify: ${\dfrac{\tan \theta \sec \theta }{\cot \theta \text{cosec}\, \theta }}$
By expressing the reciprocal functions in terms of sine and cosine, we get
${\sec \theta =\dfrac{1}{\cos \theta }}$, ${\text{cosec}\, \theta =\dfrac{1}{\sin \theta }}$, ${\tan \theta =\dfrac{\sin \theta }{\cos \theta }}$, and ${\cot \theta =\dfrac{\cos \theta }{\sin \theta }}$
Now, by substituting these values into the expression, we get
${\dfrac{\tan \theta \sec \theta }{\cot \theta \text{cosec}\, \theta }}$
= ${\dfrac{\dfrac{\sin \theta }{\cos \theta }\cdot \dfrac{1}{\cos \theta }}{\dfrac{\cos \theta }{\sin \theta }\cdot \dfrac{1}{\sin \theta }}}$
= ${\dfrac{\dfrac{\sin \theta }{\cos ^{2}\theta }}{\dfrac{\cos \theta }{\sin ^{2}\theta }}}$
= ${\dfrac{\sin \theta }{\cos ^{2}\theta }\cdot \dfrac{\sin ^{2}\theta }{\cos \theta }}$
= ${\dfrac{\sin ^{3}\theta }{\cos ^{3}\theta }}$
= ${\tan ^{3}\theta}$
Graphing
On graphing the reciprocal functions, we get
Here,
- The secant function (y = sec θ) is undefined at odd multiples of ${\dfrac{\pi }{2}}$ where cos θ = 0
- The cosecant function (y = cosec θ) is undefined at multiples of π where sin θ = 0
- The cotangent function (y = cot θ) is also undefined at multiples of π where tan θ = 0
These undefined points correspond to the vertical asymptotes of their respective graphs, as shown.
Solved Examples
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Find the values of the following reciprocal trigonometric functions:
a) ${\text{cosec}\, \dfrac{\pi }{4}}$ when ${\sin \dfrac{\pi }{4}=\dfrac{\sqrt{2}}{2}}$
b) ${\sec \dfrac{\pi }{3}}$ when ${\cos \dfrac{\pi }{3}=\dfrac{1}{2}}$
c) ${\tan \dfrac{\pi }{4}}$ when ${\cot \dfrac{\pi }{4}=1}$
Solution:
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a) As we know, sine and cosecant are reciprocal functions.
${\text{cosec}\, \theta =\dfrac{1}{\sin \theta }}$
Here, ${\text{cosec}\, \dfrac{\pi }{4}}$
= ${\dfrac{1}{\sin \dfrac{\pi }{4}}}$
= ${\dfrac{1}{\dfrac{\sqrt{2}}{2}}}$
= ${\dfrac{2}{\sqrt{2}}}$
= ${\sqrt{2}}$
Thus, ${\text{cosec}\, \dfrac{\pi }{4}}$ = ${\sqrt{2}}$
b) As we know, cosine and secant are reciprocal functions.
${\sec \theta =\dfrac{1}{\cos \theta }}$
Here, ${\sec \dfrac{\pi }{3}}$
= ${\dfrac{1}{\cos \dfrac{\pi }{3}}}$
= ${\dfrac{1}{\dfrac{1}{2}}}$
= ${2}$
Thus, ${\sec \dfrac{\pi }{3}}$ = ${2}$
c) As we know, tangent and cotangent are reciprocal functions.
${\tan \theta =\dfrac{1}{\cot \theta }}$
Here, ${\tan \dfrac{\pi }{4}}$
= ${\dfrac{1}{\cot \dfrac{\pi }{4}}}$
= ${\dfrac{1}{1}}$
= ${1}$
Thus, ${\tan \dfrac{\pi }{4}}$ = ${1}$
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Evaluate: ${\dfrac{\cos \theta }{\sin \theta }\times \text{cosec}\, \theta \times \sec \theta}$
Solution:
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Here, ${\dfrac{\cos \theta }{\sin \theta }\times \text{cosec}\, \theta \times \sec \theta}$
= ${\dfrac{\cos \theta }{\sin \theta }\times \dfrac{1}{\sin \theta }\times \dfrac{1}{\cos \theta }}$
= ${\dfrac{1}{\sin \theta }\times \dfrac{1}{\sin \theta }}$
= ${\dfrac{1}{\sin ^{2}\theta }}$
= ${\text{cosec}\, ^{2}\theta}$
Thus, ${\dfrac{\cos \theta }{\sin \theta }\times \text{cosec}\, \theta \times \sec \theta}$ = ${\text{cosec}\, ^{2}\theta}$
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Prove the identity: cosec2 θ – cot2 θ = 1
Solution:
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Here, cosec2 θ – cot2 θ
= ${\dfrac{1}{\sin ^{2}\theta }-\dfrac{\cos ^{2}\theta }{\sin ^{2}\theta }}$
= ${\dfrac{1-\cos ^{2}\theta }{\sin ^{2}\theta }}$
= ${\dfrac{\sin ^{2}\theta }{\sin ^{2}\theta }}$ (From the identity, sin2 θ + cos2 θ = 1)
= ${1}$
Thus, cosec2 θ – cot2 θ = 1, proved.
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Simplify the expression: ${\dfrac{\sec ^{2}\theta -\tan ^{2}\theta }{\text{cosec}\, ^{2}\theta -\cot ^{2}\theta }}$
Solution:
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Here, ${\dfrac{\sec ^{2}\theta -\tan ^{2}\theta }{\text{cosec}\, ^{2}\theta -\cot ^{2}\theta }}$
Using Pythagorean identities, 1 + tan2 θ = sec2 θ and 1 + cot2 θ = cosec2 θ,
= ${\dfrac{1+\tan ^{2}\theta -\tan ^{2}\theta }{1+\cot ^{2}\theta -\cot ^{2}\theta }}$
= ${\dfrac{1}{1}}$
= ${1}$
Thus, ${\dfrac{\sec ^{2}\theta -\tan ^{2}\theta }{\text{cosec}\, ^{2}\theta -\cot ^{2}\theta }}$ = ${1}$
