
What is the value of a trigonometric equation \[\left( { \dfrac{{\cot 54^{\circ }}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\tan 20^{\circ }}}{{\cot 70^{\circ }}}} \right)\] ?
A. 2
B. 3
C. 1
D. 0
Answer
218.7k+ views
Hint: First, simplify the angles in the numerator of each term as a subtraction of an angle from \[90^{\circ }\]. In the end, apply the trigonometric identities of complementary angles and simplify the equation to reach the required answer.
Formula used:
\[\tan\left( {90^{\circ } - x} \right) = \cot x\]
\[\cot\left( {90^{\circ } - x} \right) = \tan x\]
Complete step by step solution:
The given trigonometric equation is \[\left( { \dfrac{{\cot 54^{\circ }}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\tan 20^{\circ }}}{{\cot 70^{\circ }}}} \right)\].
Let’s simplify the above equation.
Let \[v\] be the value of the given trigonometric equation.
Then,
\[v = \left( { \dfrac{{\cot 54^{\circ }}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\tan 20^{\circ }}}{{\cot 70^{\circ }}}} \right)\]
Rewrite the angles in the numerator of each term as a subtraction of an angle from \[90^{\circ }\].
\[v = \left( { \dfrac{{cot \left( {90^{\circ } - 36^{\circ }} \right)}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\tan \left( {90^{\circ } - 70^{\circ }} \right)}}{{\cot 70^{\circ }}}} \right)\]
Now apply the trigonometric identities of complementary angles \[\tan\left( {90^{\circ } - x} \right) = \cot x\] and \[\cot\left( {90^{\circ } - x} \right) = \tan x\].
\[v = \left( { \dfrac{{\tan 36^{\circ}}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\cot 70^{\circ }}}{{\cot 70^{\circ }}}} \right)\]
\[ \Rightarrow \]\[v = 1 + 1\]
\[ \Rightarrow \]\[v = 2\]
Hence the correct option is A.
Note: Students often get confused about the concept of trigonometric identities of the complementary angles.
Any two angles are complementary if their sum is equal to \[90^{\circ}\]. So, the complement of any angle is the subtraction of the angle from \[90^{\circ}\].
Sine – Cosine, Tangent – Cotangent, and Secant – Cosecant are complementary angles of each other.
Following are the basic identities of the complementary angles in trigonometry:
\[\sin\left( {90^{\circ } - x} \right) = \cos x\]
\[\tan\left( {90^{\circ } - x} \right) = \cot x\]
\[\sec\left( {90^{\circ } - x} \right) = \csc x\]
Formula used:
\[\tan\left( {90^{\circ } - x} \right) = \cot x\]
\[\cot\left( {90^{\circ } - x} \right) = \tan x\]
Complete step by step solution:
The given trigonometric equation is \[\left( { \dfrac{{\cot 54^{\circ }}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\tan 20^{\circ }}}{{\cot 70^{\circ }}}} \right)\].
Let’s simplify the above equation.
Let \[v\] be the value of the given trigonometric equation.
Then,
\[v = \left( { \dfrac{{\cot 54^{\circ }}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\tan 20^{\circ }}}{{\cot 70^{\circ }}}} \right)\]
Rewrite the angles in the numerator of each term as a subtraction of an angle from \[90^{\circ }\].
\[v = \left( { \dfrac{{cot \left( {90^{\circ } - 36^{\circ }} \right)}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\tan \left( {90^{\circ } - 70^{\circ }} \right)}}{{\cot 70^{\circ }}}} \right)\]
Now apply the trigonometric identities of complementary angles \[\tan\left( {90^{\circ } - x} \right) = \cot x\] and \[\cot\left( {90^{\circ } - x} \right) = \tan x\].
\[v = \left( { \dfrac{{\tan 36^{\circ}}}{{\tan 36^{\circ }}}} \right) + \left( { \dfrac{{\cot 70^{\circ }}}{{\cot 70^{\circ }}}} \right)\]
\[ \Rightarrow \]\[v = 1 + 1\]
\[ \Rightarrow \]\[v = 2\]
Hence the correct option is A.
Note: Students often get confused about the concept of trigonometric identities of the complementary angles.
Any two angles are complementary if their sum is equal to \[90^{\circ}\]. So, the complement of any angle is the subtraction of the angle from \[90^{\circ}\].
Sine – Cosine, Tangent – Cotangent, and Secant – Cosecant are complementary angles of each other.
Following are the basic identities of the complementary angles in trigonometry:
\[\sin\left( {90^{\circ } - x} \right) = \cos x\]
\[\tan\left( {90^{\circ } - x} \right) = \cot x\]
\[\sec\left( {90^{\circ } - x} \right) = \csc x\]
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