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

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

$${\cot }^{2}x+1={\mathrm{cosec}}^{2}x$$$$\mid 4-a\mid +1=\frac{\mid a\mid }{3},0<a⩽4$$$$\Rightarrow 4-a+1=\frac{a}{3}\Rightarrow a=\frac{15}{4}$$

Commented by Rupesh123 last updated on 10/Mar/23

Excellent!

Answered by manxsol last updated on 10/Mar/23

$$sinxctx=cosx=\sqrt{\frac{3}{a}}\sqrt{4-a}$$$$\frac{3}{a}+\frac{3}{a}(4-a)=1$$$$\frac{a}{3}=5-a$$$$a=15-3a$$$$4a=15$$$$a=15/4$$$$$$

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

$$bien$$

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