Question

Prove that the following trigonometric identity is true:
  $\dfrac{{{\cot }^{2}}\theta \left( \sec \theta -1 \right)}{1+\sin \theta }={{\sec }^{2}}\theta \dfrac{1-\sin \theta }{1+\sec \theta }$

Answer 1

Hint:Take the left hand side of the equation. Try to convert the left hand side into the right hand side by using the following formulas:
 $\cot \theta =\dfrac{\cos \theta }{\sin \theta },\cos \theta =\dfrac{1}{\sec \theta },{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$

Complete step by step answer:
Let us first take the left hand side of the given equation. That is,
$\dfrac{{{\cot }^{2}}\theta \left( \sec \theta -1 \right)}{1+\sin \theta }$
Now put, $\cot \theta =\dfrac{\cos \theta }{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta }$ in the above expression.
$=\dfrac{{{\left( \dfrac{\cos \theta }{\sin \theta } \right)}^{2}}\left( \dfrac{1}{\cos \theta }-1 \right)}{1+\sin \theta }$
$=\dfrac{\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\left( \dfrac{1-\cos \theta }{\cos \theta } \right)}{1+\sin \theta }$
$=\dfrac{{{\cos }^{2}}\theta \left( 1-\cos \theta \right)}{\cos \theta {{\sin }^{2}}\theta \left( 1+\sin \theta \right)}$
Now we know that, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
We can write ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta ,{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. There we have,
$=\dfrac{\left( 1-{{\sin }^{2}}\theta \right)\left( 1-\cos \theta \right)}{\cos \theta \left( 1-{{\cos }^{2}}\theta \right)\left( 1+\sin \theta \right)}$
$=\dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)\left( 1-\cos \theta \right)}{\cos \theta \left( 1+\cos \theta \right)\left( 1-\cos \theta \right)\left( 1+\sin \theta \right)}$ , by applying the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Now we can cancel out the common terms from the numerator and the denominator.
$=\dfrac{\left( 1-\sin \theta \right)}{\cos \theta \left( 1+\cos \theta \right)}$
As the right hand side has $\sec \theta $ we will put $\cos \theta =\dfrac{1}{\sec \theta }$ in the above expression to get the required form.
$=\dfrac{1-\sin \theta }{\dfrac{1}{\sec \theta }\left( 1+\dfrac{1}{\sec \theta } \right)}$
$=\dfrac{\left( 1-\sin \theta \right)}{\dfrac{\left( 1+\sec \theta \right)}{{{\sec }^{2}}\theta }}$
$=\dfrac{{{\sec }^{2}}\theta \left( 1-\sin \theta \right)}{1+\sec \theta }$
$={{\sec }^{2}}\theta \dfrac{1-\sin \theta }{1+\sec \theta }$ , which is our right hand side.
Hence, our left hand side = our right hand side.
$\dfrac{{{\cot }^{2}}\theta \left( \sec \theta -1 \right)}{1+\sin \theta }={{\sec }^{2}}\theta \dfrac{1-\sin \theta }{1+\sec \theta }$

Note: Alternatively we can start the proof from the right hand side also.
Take the right hand side,
${{\sec }^{2}}\theta \dfrac{1-\sin \theta }{1+\sec \theta }$
Let us put $\sec \theta =\dfrac{1}{\cos \theta }$ in the above expression.
$=\dfrac{\left( 1-\sin \theta \right)}{{{\cos }^{2}}\theta \left( 1+\sec \theta \right)}$
Now we can put ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $. Therefore we have,
$=\dfrac{\left( 1-\sin \theta \right)}{\left( 1-{{\sin }^{2}}\theta \right)\left( 1+\sec \theta \right)}$
$=\dfrac{\left( 1-\sin \theta \right)}{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)\left( 1+\dfrac{1}{\cos \theta } \right)}$
Now we can cancel out the common terms from the numerator and the denominator.
$=\dfrac{\cos \theta }{\left( 1+\sin \theta \right)\left( 1+\cos \theta \right)}$
Multiply both the numerator and the denominator by $\cos \theta \left( 1-\cos \theta \right)$.
$=\dfrac{{{\cos }^{2}}\theta \left( 1-\cos \theta \right)}{\left( 1+\sin \theta \right)\left( 1-{{\cos }^{2}}\theta \right)\cos \theta }$, by putting $\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)=1-{{\cos }^{2}}\theta $
$=\dfrac{{{\cos }^{2}}\theta \left( 1-\cos \theta \right)}{{{\sin }^{2}}\theta \cos \theta \left( 1+\sin \theta \right)}$ , by putting $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $
$=\dfrac{{{\cot }^{2}}\theta \left( \dfrac{1}{\cos \theta }-1 \right)}{\left( 1+\sin \theta \right)}$ , by putting $\dfrac{1}{\cos \theta }=\sec \theta $
$=\dfrac{{{\cot }^{2}}\theta \left( \sec \theta -1 \right)}{\left( 1+\sin \theta \right)}$
Therefore the right hand side = left hand side.