if cos theta equals to [tex] \sqrt{21} \div 5[/tex]. find the value of [tex]10 + 10 \ {cot}^{2} \: theta[/tex] About the author Sophia
Answer: Step-by-step explanation: [tex]\cos \theta=\frac {\sqrt{21}}{5}\\\\\\Then \ \sin\theta =\sqrt{1-\cos^2\theta} \\\\=\sqrt{1-\frac{21}{25} }\\\\=\sqrt{\frac{4}{25} }=\frac{2}{5} \\[/tex] Now [tex]10+10\cot ^2\theta\\\\=10(1+\cot^2\theta}\\\\=10 \ cosec^2 \ \theta\\\\=\frac{10}{\sin^2\theta}\\\\=\frac{10}{\frac{4}{25}} \\\\=\frac{10\times 25}{4}\\\\=\frac{125}{2}[/tex] Reply
Answer:
Step-by-step explanation:
[tex]\cos \theta=\frac {\sqrt{21}}{5}\\\\\\Then \ \sin\theta =\sqrt{1-\cos^2\theta} \\\\=\sqrt{1-\frac{21}{25} }\\\\=\sqrt{\frac{4}{25} }=\frac{2}{5} \\[/tex]
Now
[tex]10+10\cot ^2\theta\\\\=10(1+\cot^2\theta}\\\\=10 \ cosec^2 \ \theta\\\\=\frac{10}{\sin^2\theta}\\\\=\frac{10}{\frac{4}{25}} \\\\=\frac{10\times 25}{4}\\\\=\frac{125}{2}[/tex]