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प्रश्न
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
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उत्तर
`(tan^2 theta - sec^2 theta )/ (cot^2 theta - cosec^2 theta)`
=` (-1)/(-1)`
= 1
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove the following trigonometric identities.
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If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
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cosec A(1 + cos A) (cosec A – cot A) = 1
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`sinA/(1 + cosA) = cosec A - cot A`
Prove the following identities:
`1 - sin^2A/(1 + cosA) = cosA`
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Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
What is the value of (1 − cos2 θ) cosec2 θ?
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Write True' or False' and justify your answer the following :
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Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
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`
