Question

How do you verify \[\dfrac{1-\sin }{1+\sin }={{\left( \sec -\tan \right)}^{2}}\]?

Answer 1

Hint: Here we have to prove the given problem \[\dfrac{1-\sin }{1+\sin }={{\left( \sec -\tan \right)}^{2}}\]. We should know that to prove these types of problems we have to know the trigonometric rules and the equations. We also know that if we simplify the left-hand side, we can get the right-hand side. We can simplify the left-hand side with the trigonometric formulae.

Complete step by step answer:
Now we have to prove,
\[\dfrac{1-\sin }{1+\sin }={{\left( \sec -\tan \right)}^{2}}\]
We can solve the left-hand side to get the right-hand side answer,
LHS = \[\dfrac{1-\sin }{1+\sin }\]
We know that in the denominator we have positive sign, now we can multiply by the opposite sign fraction \[\dfrac{1-\sin x}{1-\sin x}\]in the LHS part.
LHS = \[\dfrac{1-\sin }{1+\sin }\times \dfrac{1-\sin }{1-\sin }\]
In the above step, we can directly multiply the numerator and use the difference of square formula in the denominator. We also know that the difference of square formula is \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\],we can use it the denominator.
LHS = \[\dfrac{1-2\sin x+{{\sin }^{2}}x}{{{1}^{2}}-{{\sin }^{2}}x}\]
We know that \[1-{{\sin }^{2}}x={{\cos }^{2}}x\] and we apply in the above denominator.
LHS = \[\dfrac{1-2\sin x+{{\sin }^{2}}x}{{{\cos }^{2}}x}\]
Now we can separate the terms with the individual denominator, we get
LHS = \[\dfrac{1}{{{\cos }^{2}}x}-\dfrac{2\sin x}{{{\cos }^{2}}x}+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\]
LHS = \[\dfrac{1}{{{\cos }^{2}}x}-\dfrac{2\sin x}{\cos x}\times \dfrac{1}{\cos x}+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\]
Now we can simplify the above step with the formula. For the first term the formula is \[{{\sec }^{2}}x=\dfrac{1}{{{\cos }^{2}}x}\]
The formula for second term is \[\dfrac{1}{\cos x}=\sec x\] and \[\dfrac{\sin x}{\cos x}=\tan x\] , the formula for the third term is\[\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\tan }^{2}}x\]
Now we can apply the above formula in the LHS.
LHS = \[\dfrac{1}{{{\cos }^{2}}x}-\dfrac{2\sin x}{\cos x}\times \dfrac{1}{\cos x}+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\]
LHS = \[{{\sec }^{2}}x-2\sec x\tan x+{{\tan }^{2}}x\]
We know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] applying this formula in the above LHS, we get
LHS = \[{{\left( \sec x-\tan x \right)}^{2}}\]
LHS = RHS.
Hence proved that \[\dfrac{1-\sin }{1+\sin }={{\left( \sec -\tan \right)}^{2}}\].

Note:
Students make mistakes at the beginning of the problem, which is multiplying the opposite sign of the denominator on the left-hand side. We should know that to solve these types of problems, we have to know basic trigonometric rules and formulae.