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In tan θ = 1, find the value of 5cot2θ + sin2θ - 1. - Mathematics

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Sum

In tan θ = 1, find the value of 5cot2θ + sin2θ - 1.

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Solution


Consider ΔABC, where ∠A = 90°

tan θ = `"Perpendicular"/"Basse" = "AB"/"AC" = 1 = (1)/(1)`

By Pythagoras theorem,
BC2
= AB2 + AC2
= 12 + 12
= 2
⇒ BC = `sqrt(2)`
Now,

cot θ = `(1)/"tan θ"` = 1

sin θ = `"AB"/"BC" = (1)/sqrt(2)`

∴ 5cot2θ + sin2θ - 1

= `5 xx (1)^2 + (1/sqrt(2))^2 - 1`

= `5 + (1)/(2) - 1`

= `4 + (1)/(2)`

= `(9)/(2)`.

Concept: Reciprocal Relations
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APPEARS IN

Frank Class 9 Maths ICSE
Chapter 26 Trigonometrical Ratios
Exercise 26.1 | Q 18
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