Question-189026

Question Number 189026 by Rupesh123 last updated on 10/Mar/23

Answered by Ar Brandon last updated on 10/Mar/23

\cot^{2} x + 1 = {cosec}^{2} x

∣ 4 - a ∣ + 1 = \frac{∣ a ∣}{3}, 0 < a ⩽ 4

\Rightarrow 4 - a + 1 = \frac{a}{3} \Rightarrow a = \frac{15}{4}

Commented by Rupesh123 last updated on 10/Mar/23

Excellent!

Answered by manxsol last updated on 10/Mar/23

s i n x c t x = c o s x = \sqrt{\frac{3}{a}} \sqrt{4 - a}

\frac{3}{a} + \frac{3}{a} (4 - a) = 1

\frac{a}{3} = 5 - a

a = 15 - 3 a

4 a = 15

a = 15 / 4

Commented by Rupesh123 last updated on 10/Mar/23

Excellent

Answered by MJS_new last updated on 11/Mar/23

\sin^{2} x = \frac{3}{a} \Rightarrow \cos^{2} x = \frac{a - 3}{a}

\Rightarrow \cot^{2} x = \frac{a - 3}{3} = 4 - a \Rightarrow a = \frac{15}{4}

Commented by manxsol last updated on 11/Mar/23

b i e n

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