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प्रश्न
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
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उत्तर
We have to prove
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Consider the LHS.
`(cosec θ − sec θ) (cot θ − tan θ) = (1/sin theta - 1/cos theta)(cos theta/sin theta - sin theta/cos theta)`
`= ((cos theta - sin theta)/(sin theta cos theta))((cos^2 theta - sin^2 theta)/(sin theta cos theta))`
`= (cos theta - sin theta)/(sin theta cos theta) ((cos theta + sin theta)(cos theta - sin theta))/(sin theta cos theta)`
`= ((cos theta + sin theta)(cos theta - sin theta)^2)/(sin^2 theta cos^2 theta)`
Now, consider the RHS.
`(cosec θ + sec θ) ( sec θ cosec θ − 2) = (1/sin theta + 1/cos theta) (1/cos theta 1/sin theta - 2)`
`= ((cos theta + sin theta)/(sin theta cos theta))((1- 2sin theta cos theta)/(sin theta cos theta))`
`= ((cos theta + sin theta))/(sin theta cos theta) ((cos^2 theta + sin^2 theta - 2 cos theta sin theta))/(sin theta cos theta)`
`= ((cos theta + sin theta)(cos theta - sin theta)^2)/(sin^2 theta cos^2 theta)`
∴ LHS = RHS
Hence proved.
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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`
