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प्रश्न
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
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उत्तर
We have,
`6 tan^2 θ-6/cos^2 θ= 6 tan^2 θ-6 sec ^2 θ`
= `-6 (sec^2θ-tan^2 θ)` ...{`sec ^2 θ-tan ^2 θ-1` }
= -6 × 1
= -6
\[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
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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`
