Solveeit Logo
Login

Question

Question: Prove that \[\dfrac{{{{\left( {1 + \sin \theta - \cos \theta } \right)}^2}}}{{{{\left( {1 + \sin \th...

Prove that (1+sin⁡θ−cos⁡θ)2(1+sin⁡θ+cos⁡θ)2=1−cos⁡θ1+cos⁡θ\dfrac{{{{\left( {1 + \sin \theta - \cos \theta } \right)}^2}}}{{{{\left( {1 + \sin \theta + \cos \theta } \right)}^2}}} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}

Explanation

Solution

The basic idea in this type of questions is to solve the both sides of the equation separately and after simplifying the both sides check for their equality, if they are equal then the equation is true otherwise not. Also after applying componendo and dividendo the equation can be simplified easily.

Complete step-by-step solution:
The given equation is
(1+sin⁡θ−cos⁡θ)2(1+sin⁡θ+cos⁡θ)2=1−cos⁡θ1+cos⁡θ\dfrac{{{{\left( {1 + \sin \theta - \cos \theta } \right)}^2}}}{{{{\left( {1 + \sin \theta + \cos \theta } \right)}^2}}} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}
From componendo and dividendo it is known that the equation of form pq=rs\dfrac{p}{q} = \dfrac{r}{s} , it can be written as
p+qp−q=r+sr−s\dfrac{{p + q}}{{p - q}} = \dfrac{{r + s}}{{r - s}}
∴\therefore After applying componendo and dividendo in the given equation it becomes,
(1+sin⁡θ−cos⁡θ)2+(1+sin⁡θ+cos⁡θ)2(1+sin⁡θ−cos⁡θ)2−(1+sin⁡θ+cos⁡θ)2=1−cos⁡θ+1+cos⁡θ1−cos⁡θ−(1+cos⁡θ)\dfrac{{{{\left( {1 + \sin \theta - \cos \theta } \right)}^2} + {{\left( {1 + \sin \theta + \cos \theta } \right)}^2}}}{{{{\left( {1 + \sin \theta - \cos \theta } \right)}^2} - {{\left( {1 + \sin \theta + \cos \theta } \right)}^2}}} = \dfrac{{1 - \cos \theta + 1 + \cos \theta }}{{1 - \cos \theta - \left( {1 + \cos \theta } \right)}}
On simplifying,
1+sin⁡2θ+2sin⁡θ+cos⁡2θ−2(1+sin⁡θ)cos⁡θ+(1+sin⁡2θ+2sin⁡θ+cos⁡2θ+2(1+sin⁡θ)cos⁡θ)1+sin⁡2θ+2sin⁡θ+cos⁡2θ−2(1+sin⁡θ)cos⁡θ−(1+sin⁡2θ+2sin⁡θ+cos⁡2θ+2(1+sin⁡θ)cos⁡θ)=1−cos⁡θ+1+cos⁡θ1−cos⁡θ−1−cos⁡θ\dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta - 2(1 + \sin \theta )\cos \theta + (1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta + 2(1 + \sin \theta )\cos \theta )}}{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta - 2(1 + \sin \theta )\cos \theta - (1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta + 2(1 + \sin \theta )\cos \theta )}} = \dfrac{{1 - \cos \theta + 1 + \cos \theta }}{{1 - \cos \theta - 1 - \cos \theta }}
⇒1+sin⁡2θ+2sin⁡θ+cos⁡2θ−2(1+sin⁡θ)cos⁡θ+1+sin⁡2θ+2sin⁡θ+cos⁡2θ+2(1+sin⁡θ)cos⁡θ1+sin⁡2θ+2sin⁡θ+cos⁡2θ−2(1+sin⁡θ)cos⁡θ−1−sin⁡2θ−2sin⁡θ−cos⁡2θ−2(1+sin⁡θ)cos⁡θ=2−2cos⁡θ\Rightarrow \dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta - 2(1 + \sin \theta )\cos \theta + 1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta + 2(1 + \sin \theta )\cos \theta }}{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta - 2(1 + \sin \theta )\cos \theta - 1 - {{\sin }^2}\theta - 2\sin \theta - {{\cos }^2}\theta - 2(1 + \sin \theta )\cos \theta }} = \dfrac{2}{{ - 2\cos \theta }}
On solving further by cancelling the same terms having different signs,
⇒1+sin⁡2θ+2sin⁡θ+cos⁡2θ+1+sin⁡2θ+2sin⁡θ+cos⁡2θ−2(1+sin⁡θ)cos⁡θ−2(1+sin⁡θ)cos⁡θ=2−2cos⁡θ\Rightarrow \dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta + 1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta }}{{ - 2(1 + \sin \theta )\cos \theta - 2(1 + \sin \theta )\cos \theta }} = \dfrac{2}{{ - 2\cos \theta }}
Dividing by 2 in numerator and denominator of RHS,
⇒1+sin⁡2θ+2sin⁡θ+cos⁡2θ+1+sin⁡2θ+2sin⁡θ+cos⁡2θ−2(1+sin⁡θ)cos⁡θ−2(1+sin⁡θ)cos⁡θ=1−cos⁡θ\Rightarrow \dfrac{{1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta + 1 + {{\sin }^2}\theta + 2\sin \theta + {{\cos }^2}\theta }}{{ - 2(1 + \sin \theta )\cos \theta - 2(1 + \sin \theta )\cos \theta }} = \dfrac{1}{{ - \cos \theta }}
On adding the common terms having same sign in numerator and denominator of LHS,
⇒2+4sin⁡θ+2cos⁡2θ+2sin⁡2θ−4(1+sin⁡θ)cos⁡θ=1−cos⁡θ\Rightarrow \dfrac{{2 + 4\sin \theta + 2{{\cos }^2}\theta + 2{{\sin }^2}\theta }}{{ - 4(1 + \sin \theta )\cos \theta }} = \dfrac{1}{{ - \cos \theta }}
The equation can be rewritten as,
⇒−2(1+2sin⁡θ+cos⁡2θ+sin⁡2θ)4(1+sin⁡θ)cos⁡θ=−1cos⁡θ\Rightarrow \dfrac{{ - 2(1 + 2\sin \theta + {{\cos }^2}\theta + {{\sin }^2}\theta )}}{{4(1 + \sin \theta )\cos \theta }} = \dfrac{{ - 1}}{{\cos \theta }}
∵ sin⁡2θ+cos⁡2θ=1\because {\text{ }}{\sin ^2}\theta + {\cos ^2}\theta = 1
∴\therefore The equation becomes,
⇒−2(1+2sin⁡θ+1)4(1+sin⁡θ)cos⁡θ=−1cos⁡θ\Rightarrow \dfrac{{ - 2(1 + 2\sin \theta + 1)}}{{4(1 + \sin \theta )\cos \theta }} = \dfrac{{ - 1}}{{\cos \theta }}
On simplifying,
⇒−4(1+sin⁡θ)4(1+sin⁡θ)cos⁡θ=−1cos⁡θ\Rightarrow \dfrac{{ - 4(1 + \sin \theta )}}{{4(1 + \sin \theta )\cos \theta }} = \dfrac{{ - 1}}{{\cos \theta }}
On multiplying both sides of equation by −cos⁡θ- \cos \theta ,
⇒4(1+sin⁡θ)4(1+sin⁡θ)=1\Rightarrow \dfrac{{4(1 + \sin \theta )}}{{4(1 + \sin \theta )}} = 1
Dividing by 4(1+sin⁡θ)4(1 + \sin \theta ) in numerator and denominator of LHS,
⇒1=1\Rightarrow 1 = 1
∵\because After simplifying the given equation finally it comes LHS=RHSLHS = RHS ,
$\therefore $$$\dfrac{{{{\left( {1 + \sin \theta - \cos \theta } \right)}^2}}}{{{{\left( {1 + \sin \theta + \cos \theta } \right)}^2}}} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}$$

Note: The above question can also be solved by solving the both sides separately. The correctness of the equation can be checked by putting the value of theta to some angle and then solve for that. The identities must be known to solve the problem. Always try to solve the problem step by step. The calculations should be done nicely to avoid any mistake. Like componendo and dividendo other operations like invertendo, componendo, dividendo and addendo also must be known to solve this type of problems.