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Question
In an isosceles triangle ABC, AB = BC = 6 cm and ∠B = 90°. Find the values of cos2 C + cosec2 C
Solution
ΔABC is an isosceles right-angled triangle.
∴ AC2
= AB2 + BC2
= 62 + 62
= 36 + 36
= 72
⇒ AC = `6sqrt(2)"cm"`
cos2 C + cosec2 C
= `(1/sqrt(2))^2 + (sqrt(2))^2`
= `(1)/(2) + 2`
= `(5)/(2)`.
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