The general solutions of - satisfying the equation tan 2- 2 -1 can be where n - Z A- n -B- n -3C- n -4D- n -2

The correct options are A n π B n π±π3tan^2 θ + 2θ = 1 Let t = tanθ t^2 + 1+t^21 t^2 = 1 ⇒ t^2 (t^2 3) = 0 ⇒tanθ = 0 , ±√(3) ⇒θ = nπ or θ = nπ±π3, n ∈ nbsp ...

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Standard XII

Mathematics

Solving Simultaneous Trigonometric Equations

The general s...

Question

The general solution(s) of $θ$ satisfying the equation ${tan}^{2} θ + sec 2 θ = 1$ can be (where $n \in Z$ )

n π

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n π \pm \frac{π}{3}

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n π \pm \frac{π}{4}

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\frac{n π}{2}

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Solution

The correct options are
A $n π$
B $n π \pm \frac{π}{3}$
${tan}^{2} θ + sec 2 θ = 1$
Let $t = tan θ$
$t^{2} + \frac{1 + t^{2}}{1 - t^{2}} = 1$
$\Rightarrow t^{2} (t^{2} - 3) = 0 \Rightarrow tan θ = 0, \pm \sqrt{3} \Rightarrow θ = n π or θ = n π \pm \frac{π}{3}, n \in Z .$

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The general solution(s) of θ satisfying the equation tan2θ+sec2θ=1 can be (where n∈Z)

The correct options are A nπ B nπ±π3tan2θ+sec2θ=1 Let t=tanθ t2+1+t21−t2=1 ⇒t2(t2−3)=0⇒tanθ=0,±√3⇒θ=nπ or θ=nπ±π3,n∈Z.

The general solution(s) of $θ$ satisfying the equation ${tan}^{2} θ + sec 2 θ = 1$ can be (where $n \in Z$ )

The correct options are
A $n π$
B $n π \pm \frac{π}{3}$
${tan}^{2} θ + sec 2 θ = 1$
Let $t = tan θ$
$t^{2} + \frac{1 + t^{2}}{1 - t^{2}} = 1$
$\Rightarrow t^{2} (t^{2} - 3) = 0 \Rightarrow tan θ = 0, \pm \sqrt{3} \Rightarrow θ = n π or θ = n π \pm \frac{π}{3}, n \in Z .$