# Prove $\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$

$$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$$

Prove that L.H.S.$=$R.H.S.

This type of questions always creates problem when in right hand side some trigonometry function is given then it is bit easy to think how to proceed further.

Can some one help me not only to solve the problem but also how to tackle this type of other problem (when R.H.S. is $1$)?

• First try converting everything into $\sin$ and $\cos$, and then try to simplify! Jul 25, 2014 at 15:04

$$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$$

lets look at $$\frac{\cos A}{\csc A+\cot A-1} = \frac{\cos A}{\sec A+1 - \tan A}\frac{1}{\cot A} = \frac{\sin A}{\sec A+1 - \tan A}$$ this leads to

$$\frac{\sin A}{\sec A+\tan A-1}+\frac{\sin A}{\sec A+1 - \tan A} = \sin A\left[\frac{\sec A +(1-\tan A) + \sec A - (1-\tan A)}{\sec^2 A - (1-\tan A)^2}\right] = \sin A\left[\frac{2\sec A}{\sec^2 A -1-\tan ^2 A+2\tan A}\right] = \sin A\left[\frac{2\sec A }{0 + 2 \tan A}\right] = \frac{2\tan A}{2 \tan A} = 1.$$

HINT
Simplify as follows : substitute $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$. By this, we will get first term of LHS is $\frac{\sin \theta \ \cos \theta }{1 + \sin \theta - \cos \theta }$. Similar if we substitute \second term of LHS, we will get LHS =

$\frac{\sin \theta \ \cos \theta }{1 + \sin \theta - \cos \theta }$ + $\frac{\sin \theta \ \cos \theta }{1 + \cos \theta - \sin \theta }$
$= \sin \theta \ \cos\theta [\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 + \cos \theta - \sin \theta } ]$
$=\sin \theta \ \cos\theta [\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 - (\sin \theta - \cos \theta) } ]$

Now simplifying $[\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 - (\sin \theta - \cos \theta) } ]$ we get $\frac {1}{\cos \theta \sin \theta}$. But it is with product of $\sin \theta \cos \theta$. Hence the answer is $1$, Which is LHS. Hence proved.

And we should tackle these problems always by substituting in $\sin \theta$ and $\cos \theta$ when LHS is a number.

Multiply the first fraction by $\dfrac{\cos{A}}{\cos{A}}$, and the second by $\dfrac{\sin{A}}{\sin{A}}$. We get \begin{align} LHS &=\frac{\sin A\cos A}{1+\sin A- \cos A}+\frac{\sin A\cos A}{1-(\sin A- \cos A)}\\ &=\frac{\sin A\cos A-\sin A\cos A(\sin A- \cos A)+\sin A\cos A+\sin A\cos A(\sin A- \cos A)}{1-(1-2\sin A\cos A)}\\ &=\frac{2\sin A\cos A}{2\sin A\cos A}\\ &=1\\ &=RHS \end{align}

First, lets express the Pythagorean identity in another way. $$\sin^2 A +\cos^2 A = 1$$ $$\sec^2A\csc^2A(\sin^2 A +\cos^2 A) = \sec^2A\csc^2A\cdot1$$ $$\sec^2A+\csc^2 A = \sec^2A\csc^2A$$ Continuing $\dots$

$$\begin{array}{lll} \frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1} & = & \frac{1}{\csc A(\sec A+\tan A-1)}+\frac{1}{\sec A(\csc A+\cot A-1)} \\ & = & \frac{1}{\csc A\sec A+\sec A-\csc A}+\frac{1}{\sec A\csc A+\csc A-\sec A} \\ & = & \frac{1}{\sec A\csc A+(\sec A-\csc A)}+\frac{1}{\sec A\csc A-(\sec A-\csc A)} \\ & = & \frac{\sec A\csc A\color{blue}{+(\sec A-\csc A)} + \sec A\csc A\color{blue}{-(\sec A-\csc A)}}{(\sec A\csc A+(\sec A-\csc A))(\sec A\csc A-(\sec A-\csc A))}\\ & = & \frac{2\sec A\csc A}{\sec^2A \csc^2 A-(\sec A - \csc A)^2}\\ & = & \frac{2\sec A\csc A}{\sec^2A \csc^2 A-(\color{blue}{\sec^2 A + \csc^2 A} - 2\sec A\csc A)}\\ & = & \frac{2\sec A\csc A}{\sec^2A \csc^2 A-\color{blue}{\sec^2 A\csc^2 A} + 2\sec A\csc A}\\ & = & \frac{2\sec A\csc A}{2\sec A\csc A}\\ & = & 1 \end{array}$$

$$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}= \frac{\sin A}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}-\frac{\cos A}{\cos A}}+ \frac{\cos A}{\frac{1}{\sin A}+\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}} =$$ $$= \frac{\sin A \cos A}{1 + \sin A - \cos A} + \frac{\sin A \cos A}{1 + \cos A - \sin A} =$$ $$= \sin A\cos A\frac{1 + \sin A - \cos A + 1 + \cos A - \sin A}{(1+\sin A - \cos A)(1 + \cos A - \sin A)} =$$ $$=\frac{2\sin A \cos A}{1 + \cos A - \sin A + \sin A + \sin A\cos A -\sin^2 A -\cos A -\cos^2 A +\sin A\cos A} =$$ $$=\frac{2\sin A \cos A}{2\sin A\cos A} = 1$$

This type of problems can be solved in the following general way - find the derivative ( with respect to A in our case) of the left side and show that it is zero. Thus the left part is a constant. Then find this constant by putting the variable some value, for example $$A =\pi/4.$$

Expressing everything according to $$sin A$$ and $$cos A$$, we have:

$$\frac{sin A cos A}{1+sin A –cos A}+\frac{sin A cos A}{1-sin A+cos A}=1$$,

$$\frac{sin(2A)}{(1+\sqrt{2}cos(A+\frac{\pi}{4}))(1-\sqrt{2}cos(A+\frac{\pi}{4}))}=1$$,

$$\frac{sin(2A)}{1-2cos(A+\frac{\pi}{4})}=\frac{sin(2A)}{1-2(\frac{1}{2}-sin A cos A)}=$$

$$=\frac{sin(2A)}{1-(1-sin A cos A)}=\frac{sin(2A)}{2sin A cos A}=1$$.