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Question
If sin θ = `11/61`, find the values of cos θ using trigonometric identity.
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Solution
sin θ = `11/61` ...[Given]
We have,
sin2θ + cos2θ = 1
⇒ cos2θ = 1 − sin2θ
⇒ cos2θ = `1 - (11/61)^2`
⇒ cos2θ = `1 - 121/3721`
⇒ cos2θ = `(3721 - 121)/3721`
⇒ cos2θ = `3600/3721`
⇒ cos θ = `sqrt((60/61)^2)` ...[Taking the square root of both sides]
⇒ cos θ = `60/61`
Thus, the value of cos θ is `60/61`.
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sec2θ – tan2θ = ?
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
(sec2 θ – 1) (cosec2 θ – 1) is equal to ______.
Prove that (sec θ + tan θ) (1 – sin θ) = cos θ
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
