Question

Prove that $\dfrac{\operatorname{sinA}-\operatorname{cosA}+1}{\operatorname{sinA}+\cos A-1}=\dfrac{1}{\sec A-\tan A}$

Answer 1

Hint: In question it is asked that, we have to find the value of $\sin \theta $ given that $\dfrac{\operatorname{sinA}-\operatorname{cosA}+1}{\operatorname{sinA}+\cos A-1}=\dfrac{1}{\sec A-\tan A}$.
So, to do so we will use identities and properties of trigonometric ratios such as $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$,$\sin \theta =\dfrac{1}{\operatorname{cosec}\theta }$ so as to prove LHS = RHS .

Complete step by step answer:
We know that \[sin\text{ }\theta \text{ },\text{ }cos\text{ }\theta \text{ },\text{ }tan\text{ }\theta \text{ },\text{ }cot\text{ }\theta \text{ },\text{ }sec\text{ }\theta \text{ }and\text{ }cosec\text{ }\theta \] are trigonometric function, where \[\theta \] is the angle made by the hypotenuse with the base of triangle.
In question, it is given that $\dfrac{\operatorname{sinA}-\operatorname{cosA}+1}{\operatorname{sinA}+\cos A-1}=\dfrac{1}{\sec A-\tan A}$.
So, first step to solve this question which will simplify the question, will be, we will rationalise the left hand side of $\dfrac{\operatorname{sinA}-\operatorname{cosA}+1}{\operatorname{sinA}+\cos A-1}=\dfrac{1}{\sec A-\tan A}$, by ( sinA + cosA + 1 ).
Rationalization in mathematics means multiplying numerator and denominator of fraction by same factor
So, on factorization we get
LHS = $\dfrac{(\operatorname{sinA}-\operatorname{cosA}+1)(\operatorname{sinA}+\cos A+1)}{(\operatorname{sinA}+\cos A-1)(\operatorname{sinA}+\cos A+1)}$
$=\dfrac{(\operatorname{sinA}-\operatorname{cosA}+1)(\operatorname{sinA}+\cos A+1)}{(\operatorname{sinA}+\cos A-1)(\operatorname{sinA}+\cos A+1)}$
On re-arranging terms we get
$=\dfrac{[(\operatorname{sinA}+1)-\operatorname{cosA}][(\operatorname{sinA}+1)+\operatorname{cosA}]}{[(\operatorname{sinA}+\cos A)-1][(\operatorname{sinA}+\cos A)+1]}$
Now, let sinA + 1 = a and cosA = b in numerator, so we can use algebraic identity $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ in numerator.
So, we get $=\dfrac{{{(\operatorname{sinA}+1)}^{2}}-{{\cos }^{2}}A}{[(\operatorname{sinA}+\cos A)-1][(\operatorname{sinA}+\cos A)+1]}$
Now, let sinA + cosA = a and 1 = b in denominator, so we can use algebraic identity $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ in denominator.
So, we get $=\dfrac{{{(\operatorname{sinA}+1)}^{2}}-{{\cos }^{2}}A}{{{(\operatorname{sinA}+\cos A)}^{2}}-{{1}^{2}}}$
Applying algebraic identity ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ in numerator and denominator, we get
$=\dfrac{({{\sin }^{2}}A+1+2\sin A)-{{\cos }^{2}}A}{({{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A)-1}$
We know that ${{\cos }^{2}}A=1-si{{n}^{2}}A$, so we get
$=\dfrac{({{\sin }^{2}}A+1+2\sin A)-\left( 1-si{{n}^{2}}A \right)}{({{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A)-1}$
On simplifying we get
$=\dfrac{({{\sin }^{2}}A+1+2\sin A)-1+si{{n}^{2}}A}{({{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A)-1}$
Also, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$, so we get
$=\dfrac{({{\sin }^{2}}A+1+2\sin A)-1+si{{n}^{2}}A}{(1+2\sin A\cos A)-1}$
On simplifying, we get
$=\dfrac{2{{\sin }^{2}}A+2\sin A}{2\sin A\cos A}$
Taking factor 2sinA common from numerator, we get
$=\dfrac{2sinA(\operatorname{sinA}+1)}{2\sin A\cos A}$
On simplifying, we get
$=\dfrac{(\operatorname{sinA}+1)}{\cos A}$
Or, $=\dfrac{\operatorname{sinA}}{\cos A}+\dfrac{1}{\cos A}$
As we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$
So, $=\operatorname{tanA}+secA$
Multiplying numerator and denominator by $\operatorname{tanA}-secA$, we get
$=\dfrac{\left( \operatorname{tanA}+secA \right)\left( \operatorname{tanA}-secA \right)}{\operatorname{tanA}-secA}$
Using identity, $(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$, we get
$=\dfrac{\left( {{\tan }^{2}}A-se{{c}^{2}}A \right)}{\operatorname{tanA}-secA}$,
we know that ${{\tan }^{2}}A-se{{c}^{2}}A=-1$
so, we get $=\dfrac{\left( -1 \right)}{\operatorname{tanA}-secA}$
or, $=\dfrac{1}{secA-\operatorname{tanA}}$
= RHS.

Note:
One must know all trigonometric identities, properties, and the relation between trigonometric functions. While solving the question always use the most appropriate substitution of trigonometric relation which directly leads to a result. There may be calculation mistakes in cross multiplication, so be careful while solving an expression.