The general solution of nbsp- -tan-2 is-Here n- nbsp-

Given:θ+tanθ=2 ⇒1tanθ+tanθ=2 ⇒1+tan^2θtanθ=2 ⇒2tanθ1+tan^2θ=1 ⇒sin2θ=1 ⇒sin2θ=sinπ2 ⇒ 2θ=nπ+( 1)^nπ2 θ=nπ2+( 1)^nπ4 Hence, Option B is correct answer ...

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Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

The general s...

Question

The general solution of $cot θ + tan θ = 2$ is

θ = \frac{n π}{2} + (- 1)^{n} \frac{π}{8}

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θ = \frac{n π}{2} + (- 1)^{n} \frac{π}{4}

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θ = \frac{n π}{2} + (- 1)^{n} \frac{π}{6}

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θ = n π + (- 1)^{n} \frac{π}{8}

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Solution

The correct option is B $θ = \frac{n π}{2} + (- 1)^{n} \frac{π}{4}$
$cot θ + tan θ = 2 \frac{1}{tan θ} + tan θ = 2 sin 2 θ = 1 2 θ = n π + (- 1)^{n} \frac{π}{2} θ = \frac{n π}{2} + (- 1)^{n} \frac{π}{4}$

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The general solution of cotθ+tanθ=2 is

The correct option is B θ=nπ2+(−1)nπ4cotθ+tanθ=21tanθ+tanθ=2sin2θ=12θ=nπ+(−1)nπ2θ=nπ2+(−1)nπ4

The general solution of $cot θ + tan θ = 2$ is

The correct option is B $θ = \frac{n π}{2} + (- 1)^{n} \frac{π}{4}$
$cot θ + tan θ = 2 \frac{1}{tan θ} + tan θ = 2 sin 2 θ = 1 2 θ = n π + (- 1)^{n} \frac{π}{2} θ = \frac{n π}{2} + (- 1)^{n} \frac{π}{4}$