Question

If cota = 2, find all the values of all T-
ratio ofQ​

Answer 1

cot A = 2

⇒the adjacent side of A = 2

⇒the opposite side of A = 1

⇒the hypotenuse is √5

sinA = 1/√5

cos A = 2/√5

tan A = 1/2

sec A = √5/2

cosecA = √5

Cot A = 2

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Answer 2

Answer:

The answer will be :

Sin A = $\frac{1}{\sqrt{5}}$ ;

Cos A = 2/ $\sqrt{5}$;

Tan A = 1/2 ;

Cot A = 2 ;

Cosec A = $\sqrt{5}$ ;

Sec A =  $\frac{\sqrt{5} }{2}$;

Step-by-step explanation:

We have given;

                          cot A = 2;

Now, cot A = 1 / tan A;

Hence, 1 / tan A = 2;

            tan A = 1 / 2;

Now, $sec^{2}$ A- $tan^{2}$ A = 1;                    (an identity)

Hence, $sec^{2}$ A - (1/2)^2 = 1;

            $sec^{2}$ A = 1 + 1/4;

            $sec^{2}$ A = 5/4;

            sec A = $\frac{\sqrt{5} }{2}$;

Now, cos A = 1 / sec A;

Hence, cos A = 1 / ( $\frac{\sqrt{5} }{2}$);

                        = 2/ $\sqrt{5}$;

Now, $sin^{2}$ A + $cos^{2}$ A = 1;                    (an identity)

         $sin^{2}$ A + (2 / $\sqrt{5}$)^2 = 1;

         $sin^{2}$ A = 1 - 4 / 5 ;

                    = (5-4)/5;

                    = 1/5;

          sin A = $\sqrt{\frac{1}{5} }$ ;

                     = $\frac{1}{\sqrt{5}}$ ;

Now, $csc^{2}$ A = 1 /sin A;

          $csc^{2}$ A = 1 /($\frac{1}{\sqrt{5}}$);

                     = $\sqrt{5}$

           

Hence,

Sin A = $\frac{1}{\sqrt{5}}$ ;

Cos A = 2/ $\sqrt{5}$;

Tan A = 1/2 ;

Cot A = 2 ;

Cosec A = $\sqrt{5}$ ;

Sec A =  $\frac{\sqrt{5} }{2}$;

That's all.