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Find the most general value of θ satisfying the equation tan θ = –1 and cos θ = 12. - Mathematics

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प्रश्न

Find the most general value of θ satisfying the equation tan θ = –1 and cos θ = `1/sqrt(2)`.

योग
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उत्तर

Given that: tan θ = –1 and cos θ = `1/sqrt(2)`

tanθ = –1

⇒ tan θ = `tan((-pi)/4)`

⇒ tan θ = `tan(2pi - pi/4)`

⇒ tan θ = `tan  (7pi)/4`

∴ θ = `(7pi)/4`

Now cos θ = `1/sqrt(2)`

⇒ cos θ = `cos  pi/4`

⇒ cos θ = `cos(2pi - pi/4)`

⇒ cos θ = `cos  (7pi)/4`

∴ θ = `(7pi)/4`  ........[tan θ and cos θ are positive in 4th quadrant]

Hence, the most general value of θ = `2"n"pi + (7pi)/4`.

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अध्याय 3: Trigonometric Functions - Exercise [पृष्ठ ५३]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 3 Trigonometric Functions
Exercise | Q 16 | पृष्ठ ५३

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