
Find the value of trigonometric equation ${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = ?
$
{\text{A}}{\text{. }}\dfrac{1}{2}{\text{cosec 38}}^\circ {\text{ sec 22}}^\circ \\
{\text{B}}{\text{. 2 sin 22}}^\circ {\text{ cos 38}}^\circ \\
{\text{C}}{\text{. - }}\dfrac{1}{2}{\text{cosec 22}}^\circ {\text{ sec 38}}^\circ \\
{\text{D}}{\text{. none of these}} \\
$
Answer
603.9k+ views
Hint: To solve the trigonometric equation given in the question we convert the tan and cot functions in terms of sin and cos functions and simplify using basic trigonometric identities for answer.
Complete step-by-step answer:
${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{sin 38}}^\circ }}{{\cos 38^\circ }}$ - $\dfrac{{{\text{cos 22}}^\circ }}{{\sin 22^\circ }}$ -- (tanθ = $\dfrac{{\sin \theta }}{{\cos \theta }}$ and cotθ =$\dfrac{{\cos \theta }}{{\sin \theta }}$)
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{sin 38}}^\circ {\text{sin 22}}^\circ - \cos {\text{ 22}}^\circ \cos {\text{ 38}}^\circ }}{{\cos 38^\circ {\text{sin 22}}^\circ }}$ --- (1)
We know that,
cos (A + B) = cos A cos B – sin A sin B
Using the above equation in equation (1), we get
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{ - cos}}\left( {38 + 22} \right)^\circ }}{{\cos 38^\circ {\text{sin 22}}^\circ }}$
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{ - cos}}\left( {60} \right)^\circ }}{{\cos 38^\circ {\text{sin 22}}^\circ }}$
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{ - }}\dfrac{1}{2}}}{{\cos 38^\circ {\text{sin 22}}^\circ }}$
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $ - \dfrac{1}{2}{\text{cosec 22}}^\circ {\text{ sec 38}}^\circ $
Hence, Option C is the correct answer.
Note: In order to solve this type of questions the key is to convert the given trigonometric functions into a different form such that it can be simplified using a known trigonometric identity. Then we reduce it into another simple trigonometric value or a function using appropriate formulae, to determine the answer. Basic knowledge of trigonometric functions and identities is essential.
Complete step-by-step answer:
${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{sin 38}}^\circ }}{{\cos 38^\circ }}$ - $\dfrac{{{\text{cos 22}}^\circ }}{{\sin 22^\circ }}$ -- (tanθ = $\dfrac{{\sin \theta }}{{\cos \theta }}$ and cotθ =$\dfrac{{\cos \theta }}{{\sin \theta }}$)
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{sin 38}}^\circ {\text{sin 22}}^\circ - \cos {\text{ 22}}^\circ \cos {\text{ 38}}^\circ }}{{\cos 38^\circ {\text{sin 22}}^\circ }}$ --- (1)
We know that,
cos (A + B) = cos A cos B – sin A sin B
Using the above equation in equation (1), we get
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{ - cos}}\left( {38 + 22} \right)^\circ }}{{\cos 38^\circ {\text{sin 22}}^\circ }}$
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{ - cos}}\left( {60} \right)^\circ }}{{\cos 38^\circ {\text{sin 22}}^\circ }}$
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $\dfrac{{{\text{ - }}\dfrac{1}{2}}}{{\cos 38^\circ {\text{sin 22}}^\circ }}$
⟹${\text{tan38}}^\circ - {\text{cot22}}^\circ $ = $ - \dfrac{1}{2}{\text{cosec 22}}^\circ {\text{ sec 38}}^\circ $
Hence, Option C is the correct answer.
Note: In order to solve this type of questions the key is to convert the given trigonometric functions into a different form such that it can be simplified using a known trigonometric identity. Then we reduce it into another simple trigonometric value or a function using appropriate formulae, to determine the answer. Basic knowledge of trigonometric functions and identities is essential.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

