Question

If tan theta= 2 root 2 then find other 5 trigonometry ratioes

Answer 1

Step-by-step explanation:

ANSWER IS GIVEN BELOW

$\tan(a) = 2 \sqrt{2} \\ \\ \tan(a) = \frac{1}{ \cot(a) } \\ \\ ((( \cot(a) = \frac{1}{2 \sqrt{2} } ))) \\ \\ \\ \tan(a) = \frac{ \sin(a) }{ \cos(a) } \\ \\ 2 \sqrt{2} = \frac{ \sin(a) }{ \cos(a) } \\ \\ = ((\sin(a) = 2 \sqrt{2} \cos(a) )))$

Answer 2

Answer:

• Sin (theta) = $\frac{2\sqrt{2} }{3}$
• Cos (theta) = (1/3)
• Cot (theta) = $\frac{1\\}{2\sqrt{2} }$
• Cosec (theta) = $\frac{3}{2\sqrt{2} }$
• Sec (theta) = 3

Step-by-step explanation:

Given, Tan (Theta) = $2\sqrt{2}$

To Find: The values of all other 5 trigonometric ratios.

Explanation:

• For the ease of answering and depiction, let us assume "theta" = x.

        Then, Tan (x) = $2\sqrt{2$

• Now to find the values of all other 5 trigonometric ratios, we will need to know about the trigonometric identity equations.
• First is [{Tan (x)} * {Cot (x)}] = 1,

        $or, Cot(x)=\frac{1}{Tan(x)} \\or, Cot (x) = \frac{1}{2\sqrt{2} }$

• Second is, [{Tan (x)}^2] + 1 =  [{Sec (x)}^2],

       $or, [1 + Tan^{2}(x)]=Sec^{2}(x)\\or, \sqrt{Sec^{2}(x)} = \sqrt{[1 + Tan^{2}(x)]} \\or, Sec(x)=\sqrt{1+(2\sqrt{2}) ^{2} }\\ or,Sec(x)=\sqrt{1+8}\\ or,Sec(x)=\sqrt{9}\\ or, Sec(x) = 3$

• Then, [{Cos (x)} * {Sec (x)}] = 1,

        $or, Cos(x)=\frac{1}{Sec(x)} \\or, Cos (x) = \frac{1}{3}$

• Now, [{Sin (x)}^2] + [{Cos (x)}^2] =  1,

        $or, [Sin^{2}(x)+Cos^{2}(x)}=1\\or,Cos^{2}(x)=[1-Sin^{2}(x)]\\ \sqrt{Sin^{2}(x)} = \sqrt{[1 - Cos^{2}(x)]} \\or, Sin(x)=\sqrt{1-(\frac{1}{3} ) ^{2} }\\ or,Sin(x)=\sqrt{1-\frac{1}{9} }\\ or,Sin(x)=\sqrt{\frac{9-1}{9} }\\ or, Sin(x) = \sqrt{\frac{8}{9} } \\or, Sin(x)=\frac{2\sqrt{2} }{3}$

• Finally, [{Sin (x)} * {Cosec (x)}] = 1,

        $or, Cosec(x)=\frac{1}{Sin(x)} \\or, Cosec (x) = \frac{1}{\frac{2\sqrt{2} }{3} }\\or, Cosec(x)=\frac{3}{2\sqrt{2} }$

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