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प्रश्न
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
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उत्तर
We have,
\[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]=` (-1(sec ^2 θ-tan ^2θ ))/(-1 (cosec^2 θ-cot ^2 θ))`
=`( secx^2θ-tan^2 θ)/ (cosec ^2 θ-cot^2 θ)`
We know that,
`sec^2θ-tan ^2θ=1`
` cosec^2 θ-cot ^2θ=1`
Therefore,
`(tan ^2θ-sec^2 θ)/(cot^2θ-cosec^2 θ)=1/1`
=1
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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`
