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प्रश्न
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
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उत्तर
L.H.S. = tan2Φ + cot2Φ + 2
= tan2Φ + 1 + cot2Φ + 1
= sec2Φ + cosec2Φ
= `1/cos^2 Φ + 1/sin^2Φ`
= `(sin^2 Φ + cos^2 Φ)/(sin^2 Φ.cos^2Φ )`
= `1/(sin^2 Φ. cos^2 Φ )`
= cosec2Φ. sec2Φ
= R.H.S.
Hence proved.
संबंधित प्रश्न
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If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
If `sec theta + tan theta = p,` prove that
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If x = a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`
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Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove that
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
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`
