Question

Prove that$\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta .\cos ec\theta $

Answer 1

Hint: In the above question we will try to solve from Right Hand Side i.e., RHS. We need to know some basic formula of trigonometry like $\tan \theta =\dfrac{{\sin \theta }}{{\cos \theta }}$, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$.

Complete step-by-step answer: Given,
$\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta .\cos ec\theta $
$ \Rightarrow\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}$ (Putting the value of tan$\theta $ and cot$\theta $ in Right Hand Side)
$ \Rightarrow\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{\dfrac{{\cos \theta - \sin \theta }}{{\cos \theta }}}}$(Taking LCM)
$ \Rightarrow\dfrac{{{{\sin }^2}\theta }}{{\cos \theta (\sin \theta - \cos \theta )}} + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta (\cos \theta - \sin \theta )}}$
$ \Rightarrow\dfrac{{{{\sin }^2}\theta }}{{\cos \theta (\sin \theta - \cos \theta )}} - \dfrac{{{{\cos }^2}\theta }}{{\sin \theta (\sin \theta - \cos \theta )}}$(Taking ‘- ‘common)
$ \Rightarrow\dfrac{{{{\sin }^3}\theta - {{\cos }^3}\theta }}{{\cos \theta \sin \theta (\sin \theta - \cos \theta )}}$Taking LCM)
$ \Rightarrow\dfrac{{\left( {\sin \theta - \cos \theta } \right)({{\sin }^2}\theta + \sin \theta .\cos \theta + {{\cos }^2}\theta )}}{{\cos \theta \sin \theta (\sin \theta - \cos \theta )}}$(Using formula ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
$ \Rightarrow\dfrac{{{{\sin }^2}\theta + \sin \theta .\cos \theta + {{\cos }^2}\theta }}{{\cos \theta .\sin \theta }}$
$ \Rightarrow\dfrac{{1 + \sin \theta .\cos \theta }}{{\cos \theta .\sin \theta }}$(we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$)
$ \Rightarrow\dfrac{1}{{\cos \theta .\sin \theta }} + \dfrac{{\sin \theta .\cos \theta }}{{\cos \theta .\sin \theta }}$
$ \Rightarrow \sec \theta .\cos ec\theta + 1$
Hence, it is proved that $\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta .\cos ec\theta $.

Note: Trigonometry, the branch of mathematics concerned with specific functions of angles and their application to calculations. There are six functions of an angle commonly used in trigonometry. Their names and abbreviations are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (cosec). Trigonometric formulae are used in obtaining unknown angles and distances from known or measured angles in geometric figures.
The three basic functions in trigonometry are sine, cosine and tangent. Based on these three functions the other three functions that are cotangent, secant and cosecant are derived. All the trigonometrical concepts are based on these functions. Hence, to understand trigonometry further we need to learn these functions and their respective formulas at first.
If θ is the angle in a right-angled triangle, then
Sin θ = Perpendicular/Hypotenuse
Cos θ = Base/Hypotenuse
Tan θ = Perpendicular/Base
Perpendicular is the side opposite to the angle θ.
The base is the adjacent side to the angle θ.
The hypotenuse is the side opposite to the right angle
The other three functions i.e. cot, sec and cosec depend on tan, cos and sin respectively, such as:
Cot θ = 1/tan θ
Sec θ = 1/cos θ
Cosec θ = 1/sin θ
Hence,
Cot θ = Base/Perpendicular
Sec θ = Hypotenuse/Base
Cosec θ = Hypotenuse/Perpendicular