Question

The following equation is equals to:-
$\frac{1}{ \csc(1 - \cot ) } + \frac{1}{ \sec(1 - \tan ) }$
​

Answer 1

Before starting this question I would like to state that you should take these trigonometrical ratios with respect to an angle for ease. Here, we'll take A as respective angle.

Question

The following equation is equals to :-

${ \mathcal{\dfrac{1}{ \cosec A(1 - \cot A ) } + \dfrac{1}{ \sec A(1 - \tan A)}}}$

Answer

$= \rm( \sin A + \cos A)$

Solution

We know that sinA = 1/cosecA & cosA = 1/secA ( reciprocal relation )

$\rm \dfrac{ \sin A}{1 - \cot A} + \dfrac{ \cos A}{1 - tan A}$

Also, from Quotient Relation ,

tanA = sinA/cosA & cotA = cosA/sinA

$= \rm \dfrac{ \sin A}{1 - \dfrac{ \cos A}{ \sin A} } + \dfrac{ \cos A}{1 - \dfrac{ \sin A}{ \cos A} }$

$\rm{ = \dfrac{ \sin A}{\dfrac{ \sin A - \cos A}{ \sin A}} + \dfrac{ \cos A}{ \dfrac{ \cos A - \sin A}{ \cos A} } }$

$= \rm\dfrac{ \sin {}^{2} A}{ \sin A - cos A} + \dfrac{ \cos {}^{2} A }{ \cos A - \sin A}$

Note that in order to make a common base I.e sinA - cosA , we will rearrange the second term

$= \rm\dfrac{ \sin {}^{2} A}{ \sin A - cos A} - \dfrac{ \cos {}^{2} A }{ \sin A - \cos A}$

$= \rm\dfrac{ \sin {}^{2} A - \cos {}^{2}A }{ \sin A - \cos A}$

$= \rm\dfrac{ \cancel{(\sin A - \cos A})(\sin A + \cos A) }{ \cancel{\sin A - \cos A}}$

$= \rm( \sin A + \cos A)$

Hence , we get the required value obtained after simplifying the above trigonometrical equations

$\underline{\rule{190pt}{2pt}}$

Thankyou