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Find the value of the trigonometric functions for the following:tan θ = −2, θ lies in the II quadrant

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प्रश्न

Find the value of the trigonometric functions for the following:
tan θ = −2, θ lies in the II quadrant

बेरीज
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उत्तर

We know that sec2θ – tan2θ = 1

sec2θ – (– 2)2 = 1

sec2θ – 4 = 1

sec2θ = 1 + 4 = 5

sec θ = `+-  sqrt(5)`

Since θ lies in the second quadrant sec θ is negative.

∴ sec θ = `- sqrt(5)`

cos θ = `1/sectheta = -1/sqrt(5)`

We know cos2θ + sin2θ = 1

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

`1/5 + sin^2theta` = 1

sin2θ = `1 - 1/5 = (5 - 1)/5`

sin2θ = `4/5`

sin θ = `+-  2/sqrt(5)`

Since θ lies in the second quadrant sin θ is positivee.

∴ sin θ = `2/sqrt(5)`

sin θ =  `2/sqrt(5)`, cosec  = `1/sintheta = sqrt(5)/2`

cos θ = `- 1/sqrt(5)`, sec θ = `1/costheta = - sqrt(5)`

tan θ =  – 2, cot θ = `1/tantheta = - 1/2`

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Trigonometric Functions and Their Properties
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.3 [पृष्ठ १०४]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.3 | Q 3. (iv) | पृष्ठ १०४

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