Question

How do you prove $\dfrac{{\csc B}}{{\cos B}} - \dfrac{{\cos B}}{{\sin B}} = \tan B$

Answer 1

To prove the given trigonometric function, apply trigonometric identity formulas i.e., applying reciprocal formula of cosec in terms of sin function, as we know that sec, csc and cot are derived from primary functions of sin, cos and tan, hence using these functions we can prove that LHS = RHS of the given function.

Complete step-by-step solution:
The given function is
$\Rightarrow \dfrac{{\csc B}}{{\cos B}} - \dfrac{{\cos B}}{{\sin B}} = \tan B$
Let us consider the LHS term i.e.,
$\Rightarrow \dfrac{{\csc B}}{{\cos B}} - \dfrac{{\cos B}}{{\sin B}}$
Apply the reciprocal formula of cosec in the equation we get
$\Rightarrow \dfrac{{\dfrac{1}{{\sin B}}}}{{\cos B}} - \dfrac{{\cos B}}{{\sin B}}$
Simplifying the terms by dividing $\cos B$to the equation as
$\Rightarrow \dfrac{1}{{\sin B}}\times \dfrac{1}{{\cos B}} - \dfrac{{\cos B}}{{\sin B}}$
Multiplying the functions, we get
$\Rightarrow \dfrac{1}{{\sin B\times \cos B}} - \dfrac{{\cos B}}{{\sin B}}$
Simplifying we get
$\Rightarrow \dfrac{{1 - \cos B\left( {\cos B} \right)}}{{\sin B\times \cos B}}$
$\Rightarrow \dfrac{{1 - {{\cos }^2}B}}{{\sin B\times \cos B}}$
We know that
$\Rightarrow 1 - {\cos ^2}B$= ${\sin ^2}B$
Hence, we get
$\Rightarrow \dfrac{{si{n^2}B}}{{\sin B\times \cos B}}$
$\Rightarrow \dfrac{{\sin B\sin B}}{{\sin B\times \cos B}}$
The similar terms imply to one, we get
$\Rightarrow \dfrac{{\sin B}}{{\tan B}}$
$\Rightarrow \tan B$ = RHS

Therefore, hence proved $\dfrac{{\csc B}}{{\cos B}} - \dfrac{{\cos B}}{{\sin B}} = \tan B$

Additional Information:
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 = $\dfrac{{perpendicular}}{{hypotenuse}}$
Cos θ = $\dfrac{{base}}{{hypotenuse}}$
Tan θ = $\dfrac{{perpendicular}}{{base}}$
The other three functions i.e., cot, sec and cosec depend on tan, cos and sin respectively.

Note: The key point to prove any trigonometric function is to note the formulas of all related functions with respect to the equation, and prove all the terms by considering LHS terms and prove for RHS and to prove the given function we must know the reciprocal formula of cosec and hence, by applying this and simplifying the terms of the function we can prove LHS = RHS.