15 points for 3 question
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Answer:
These three questions can be solved by applying trigonometric identities.
Step-by-step explanation:
I am providing step by step explanation of all three questions -
Q30 - tan²A - tan²B = (cos²B - cos²A)/(cos²A * cos²B)
= (sin²A - sin²B)/(cos²A * cos²B)
Solution - LHS = tan²A - tan²B
= (sin²A/cos²A) - (sin²B/cos²B)
= (sin²A * cos²B - sin²B * cos²A)/(cos²A * cos²B)
= [sin²A * (1 - sin²B) - sin²B * (1 - sin²A)]/(cos²A * cos²B)
= (sin²A - sin²A *sin²B - sin²B + sin²A *sin²B)/(cos²A * cos²B)
= (sin²A - sin²B)/(cos²A * cos²B)
Hence Proved .
Also,LHS= [cos²B * (1 - cos²A) - cos²A * (1 - cos²B)]/(cos²A * cos²B)
= (cos²B - cos²A *cos²B - cos²A + cos²A *cos²B)/(cos²A * cos²B)
= (cos²B -cos²A)/(cos²A * cos²B)
= RHS
Hence proved .
Q31 - 1/(cosecθ + cotθ) - 1/sinθ = 1/sinθ - 1/(cosecθ - cotθ)
Solution - LHS = 1/(cosecθ + cotθ) - 1/sinθ
= [1/(1/sinθ + cosθ/sinθ)] - 1/sinθ
= sinθ/(1 + cosθ) - 1/sinθ
= (sin²θ - 1 - cosθ)/[sinθ(1 + cosθ)]
= (1 - cos²θ - 1 - cosθ)/[sinθ(1 + cosθ)]
= (- cos²θ - cosθ)/[sinθ(1 + cosθ)]
= [-cosθ(1 + cosθ)]/[sinθ(1 + cosθ)]
= - cotθ
RHS = 1/sinθ - 1/(cosecθ - cotθ)
= 1/sinθ - [1/(1/sinθ - cosθ/sinθ)]
= 1/sinθ - sinθ/(1 - cosθ)
= (- sin²θ + 1 - cosθ)/[sinθ(1 - cosθ)]
= (- 1 + cos²θ + 1 - cosθ)/[sinθ(1 - cosθ)]
= ( cos²θ - cosθ)/[sinθ(1 - cosθ)]
= [-cosθ(1 - cosθ)]/[sinθ(1 - cosθ)]
= - cotθ
Thus, LHS = RHS . Hence Proved .
Q32 - If secθ - tanθ = k
Solution - Firstly , we can write reciprocal of above equation
=> 1/(secθ - tanθ) = 1/k
Than rationalise the equation
=> (secθ + tanθ)/[(secθ + tanθ)*(secθ - tanθ)] = 1/k
=> (secθ + tanθ)/(sec²θ - tan²θ) = 1/k
=> (secθ + tanθ) = 1/k ....................(1)
Also, (secθ - tanθ) = k ....................(2)
Adding equations (1) and (2)
=> 2secθ = k + 1/k
=> 2secθ = (k² + 1)/k
=> secθ = (k² + 1)/2k
=> cosθ = 2k/(k² + 1) . Hence Proved
As, cosθ = 2k/(k² + 1) = Base/Hypotenuse
=> Perpendicular,P = √H²-B²
= √(k² + 1)² - (2k)²
= √(k² - 1)²
= k² - 1
So, sinθ = Perpendicular/Hypotenuse
= (k² - 1)/(k² + 1)
=> sinθ = (k² - 1)/(k² + 1) . Hence Proved