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Question: If \( b\cos \theta = a \) Prove that - \( \cos ec\theta + \cot \theta = \sqrt {\dfrac{{b + a}}{{b - ...

If bcos⁡θ=ab\cos \theta = a Prove that - cos⁡ecθ+cot⁡θ=b+ab−a\cos ec\theta + \cot \theta = \sqrt {\dfrac{{b + a}}{{b - a}}}

Explanation

Solution

Hint : Here, use the trigonometric functions and its simplification, then substitute the given cosine angle value and simplify using basic mathematical operations and different properties of the difference of the squares and square-roots. That is n=n×nn = \sqrt n \times \sqrt n .

Complete step-by-step answer :
bcos⁡θ=ab\cos \theta = a
Make cos⁡θ\cos \theta the subject –
cos⁡θ=ab   .......(1)\cos \theta = \dfrac{a}{b}\;{\text{ }}.......{\text{(1)}}
Using the trigonometric identity that –
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Make the subject
⇒sin⁡2θ=1−cos⁡2θ  \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta \\\ \Rightarrow sin⁡θ=1−cos⁡2θ \sin \theta = \sqrt {1 - {{\cos }^2}\theta } \\\
Place the given value in the right hand side of the equation –
⇒sin⁡θ=1−(ab)2\Rightarrow \sin \theta = \sqrt {1 - {{\left( {\dfrac{a}{b}} \right)}^2}}
Simplify the above right hand side of the equation –
⇒sin⁡θ=1−ab22\Rightarrow \sin \theta = \sqrt {1 - {{\dfrac{a}{{{b^2}}}}^2}}
Take LCM (Least common factor) on the right hand side of the equation and simplify it.

sin⁡θ=b2−a2b2 sin⁡θ=b2−a2b ........(2)  \sin \theta = \sqrt {\dfrac{{{b^2} - {a^2}}}{{{b^2}}}} \\\ \sin \theta = \dfrac{{\sqrt {{b^2} - {a^2}} }}{b}{\text{ }}........{\text{(2)}} \\\

(As, square and square-root cancel each other in the denominator)
Now, take the given Left hand side of the equation –
LHS =cos⁡ecθ+cot⁡θ= \cos ec\theta + \cot \theta
Convert the above equation in the terms of sin⁡θ and cosθ\sin \theta {\text{ and cos}}\theta , where ⇒cos⁡ecθ=1sin⁡θ  and cotθ = cos⁡θsin⁡θ\Rightarrow \cos ec\theta = \dfrac{1}{{\sin \theta }}\;{\text{and cot}}\theta {\text{ = }}\dfrac{{\cos \theta }}{{\sin \theta }}
LHS =1sin⁡θ+cos⁡θsin⁡θ= \dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}
Since, the denominator of both the terms are the same, add numerator directly.
LHS =1+cos⁡θsin⁡θ= \dfrac{{1 + \cos \theta }}{{\sin \theta }}
Substitute values from the equation (1)  and (2)(1)\;{\text{and (2)}}
LHS =1+abb2−a2b= \dfrac{{1 + \dfrac{a}{b}}}{{\dfrac{{\sqrt {{b^2} - {a^2}} }}{b}}}
Take LCM on the numerator part of the equation on the right
LHS =b+abb2−a2b= \dfrac{{\dfrac{{b + a}}{b}}}{{\dfrac{{\sqrt {{b^2} - {a^2}} }}{b}}}
Numerator’s denominator and denominator’s denominator cancel each other.
LHS =b+ab2−a2= \dfrac{{b + a}}{{\sqrt {{b^2} - {a^2}} }}
Using the property of the difference of two squares is - (b2−a2=(b+a)(b−a))(\sqrt {{b^2} - {a^2}} = \sqrt {(b + a)(b - a)} )
Also, the square is the product of its square-root into square-root, n=n×nn = \sqrt n \times \sqrt n
⇒\Rightarrow LHS =(b+a)×(b+a)(b+a)(b−a)= \dfrac{{\sqrt {(b + a)} \times \sqrt {(b + a)} }}{{\sqrt {(b + a)(b - a)} }}
⇒\Rightarrow LHS =(b+a)×(b+a)(b+a)×(b−a)= \dfrac{{\sqrt {(b + a)} \times \sqrt {(b + a)} }}{{\sqrt {(b + a)} \times \sqrt {(b - a)} }}
Same terms from the numerator and the denominator cancel each other.
⇒\Rightarrow LHS =(b+a)(b−a)= \dfrac{{\sqrt {(b + a)} }}{{\sqrt {(b - a)} }}
⇒\Rightarrow LHS =b+ab−a= \sqrt {\dfrac{{b + a}}{{b - a}}}
LHS=RHS
Hence, the given statement is proved.

Note : Remember the basic trigonometric formulas and apply them accordingly. Directly the Pythagoras identity are followed by sines and cosines which states that – sin2θ+cos2θ=1si{n^2}\theta + co{s^2}\theta = 1 and derive other trigonometric functions using it such as tan, cosec, cot and cosec angles.