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Maths NCERT Exemplar Solutions Class 12th Chapter One Permutations and Combinations Ncert Solutions Maths class 11th Maths Class 11th

Let S = ${θ \in [- π, π] - {\pm \frac{π}{2}} : s i n θ t a n θ + t a n θ = s i n 2 θ}$ . If $\sum_{θ \in S}^{} c o s 2 θ,$ then T = n(S) is equal to

Option 1 -

$7 + \sqrt[]{3}$

Option 2 -

Option 3 -

$8 + \sqrt[]{3}$

Option 4 -

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$\frac{s i n θ s i n θ}{c o s θ} + \frac{s i n θ}{c o s θ} = 2 s i n θ c o s θ$

$\Rightarrow \frac{s i n θ}{c o s θ} [s i n θ + 1] = 2 s i n θ c o s θ$

$∴$ sinθ = 0 and 1 + sin θ = 2 cos² θ = 2 – 2 sin² θ ………….(i)

θ = np

θ = -p, p, 0

From (i), 2 sin² θ + sin θ - 1 = 0

(2 sin θ - 1) (sin θ + 1) = 0

sin θ = -1, $\frac{1}{2}$

$θ = \frac{3 π}{2}, θ = \frac{π}{6}, \frac{5 π}{6}$

$θ = \frac{3 π}{2}$ is rejected.

T = cos (-2p) + cos 2p + cos θ + cos $\frac{π}{3} + c o s \frac{5 π}{3} = 4$

$∴$ T + n(s) = 4 + 5 = 9

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Let S = ${θ \in [- π, π] - {\pm \frac{π}{2}} : s i n θ t a n θ + t a n θ = s i n 2 θ}$ . If $\sum_{θ \in S}^{} c o s 2 θ,$ then T = n(S) is equal to

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Let S = { θ ∈ [ − π , π ] − { ± π 2 } : s i n θ t a n θ + t a n θ = s i n 2 θ } . If ∑ θ ∈ S c o s 2 θ , then T = n(S) is equal to

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Let S = ${θ \in [- π, π] - {\pm \frac{π}{2}} : s i n θ t a n θ + t a n θ = s i n 2 θ}$ . If $\sum_{θ \in S}^{} c o s 2 θ,$ then T = n(S) is equal to