Class 10 math question
Answers
Option (d)
Solution :-
Given that
(cos θ - sin θ + 1)/( cos θ + sin θ -1)
It can be written as
=> [cos θ-(sin θ-1)] / [cos θ+(sin θ-1)]
On multiplying both numerator and denominator with [cos θ-(sin θ-1)] then
=>[{cos θ-(sin θ-1)}/{cos θ+(sin θ-1)}]
×[{cos θ-(sin θ-1)} /{cos θ-(sin θ-1)}]
=> [{cos θ-(sin θ-1)}{cos θ-(sin θ-1)}] /
[{cos θ+(sin θ-1)}{cos θ-(sin θ-1)}]
=> [cos θ-(sin θ-1)]² / [(cos θ)²-(sin θ-1)²]
Since, (a+b)(a-b) = a²-b²
=> [cos θ-(sin θ-1)]² / [cos²θ-(sin θ-1)²]
We know that
(a-b)² = a²-2ab+b²
=> [cos² θ-2cos θ(sin θ-1)+sin² θ-2sin θ+1]/[cos²θ-(sin² θ-2sin θ+1)]
=> [cos² θ-2cos θ sin θ+2cos θ+sin² θ
-2sin θ+1] / [cos² θ-sin² θ+2sin θ-1]
=> [1-2 cos θ sin θ+2cos θ-2sin θ+1] / [cos²θ-sin²θ+2 sin θ-1]
Since , sin² A + cos² A = 1
=> [1-2 cos θ sin θ+2 cos θ-2sin θ+1] /[1-sin²θ-sin²θ+2 sin θ-1]
=>(2-2cosθ sinθ+2cosθ-2sinθ)/(-sin²θ-sin²θ+2sinθ)
=>2(1-cosθ sinθ+cosθ-sinθ)/(-2 sin² θ+2 sinθ)
=> 2(1-cosθ sinθ+cos θ- sin θ)/2(- sin² θ +sin θ)
=> (1-cos θ sin θ+cos θ- sin θ)/(- sin² θ
+sin θ)
=>[(1+cos θ)-sinθ(1+cosθ)]/[sinθ(1-sin θ)]
=> [(1+cos θ)(1-sin θ)]/[sin θ(1-sin θ)]
On cancelling (1-sin θ) in both numerator and denominator then
=> (1+cos θ)/(sin θ)
=> (1/sinθ) + (cos θ/sin θ)
=> cosec θ+ cot θ
Therefore,
(cosθ-sinθ+1)/(cosθ+sinθ-1)=cosecθ+cotθ
Used formulae:-
→ (a+b)(a-b) = a²-b²
→ (a-b)² = a²-2ab+b²
→ sin²A + cos² A = 1
Step-by-step explanation:
Option (d)
Solution :-
Given that
(cos θ - sin θ + 1)/( cos θ + sin θ -1)
It can be written as
=> [cos θ-(sin θ-1)] / [cos θ+(sin θ-1)]
On multiplying both numerator and denominator with [cos θ-(sin θ-1)] then
=>[{cos θ-(sin θ-1)}/{cos θ+(sin θ-1)}]
×[{cos θ-(sin θ-1)} /{cos θ-(sin θ-1)}]
=> [{cos θ-(sin θ-1)}{cos θ-(sin θ-1)}] /
[{cos θ+(sin θ-1)}{cos θ-(sin θ-1)}]
=> [cos θ-(sin θ-1)]² / [(cos θ)²-(sin θ-1)²]
Since, (a+b)(a-b) = a²-b²
=> [cos θ-(sin θ-1)]² / [cos²θ-(sin θ-1)²]
We know that
(a-b)² = a²-2ab+b²
=> [cos² θ-2cos θ(sin θ-1)+sin² θ-2sin θ+1]/[cos²θ-(sin² θ-2sin θ+1)]
=> [cos² θ-2cos θ sin θ+2cos θ+sin² θ
-2sin θ+1] / [cos² θ-sin² θ+2sin θ-1]
=> [1-2 cos θ sin θ+2cos θ-2sin θ+1] / [cos²θ-sin²θ+2 sin θ-1]
Since , sin² A + cos² A = 1
=> [1-2 cos θ sin θ+2 cos θ-2sin θ+1] /[1-sin²θ-sin²θ+2 sin θ-1]
=>(2-2cosθ sinθ+2cosθ-2sinθ)/(-sin²θ-sin²θ+2sinθ)
=>2(1-cosθ sinθ+cosθ-sinθ)/(-2 sin² θ+2 sinθ)
=> 2(1-cosθ sinθ+cos θ- sin θ)/2(- sin² θ +sin θ)
=> (1-cos θ sin θ+cos θ- sin θ)/(- sin² θ
+sin θ)
=>[(1+cos θ)-sinθ(1+cosθ)]/[sinθ(1-sin θ)]
=> [(1+cos θ)(1-sin θ)]/[sin θ(1-sin θ)]
On cancelling (1-sin θ) in both numerator and denominator then
=> (1+cos θ)/(sin θ)
=> (1/sinθ) + (cos θ/sin θ)
=> cosec θ+ cot θ
Therefore,
(cosθ-sinθ+1)/(cosθ+sinθ-1)=cosecθ+cotθ
Used formulae:-
→ (a+b)(a-b) = a²-b²
→ (a-b)² = a²-2ab+b²
→ sin²A + cos² A = 1