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Trigonometric Ratios CBSE NCERT Solutions Chapter 8 Exercise 8.2 Question 2

 

Trigonometric Ratios CBSE NCERT Solutions Chapter 8 Exercise 8.2 Question 2


 

2.   Choose the correct option and justify your choice:


You can check here values of six trigonometric ratios (sin, cos, tan, sec, cot and cosec) for 0, 30, 45, 60 and 90 degrees. You can also learn from this article about how to memorize all the values in an easy way.

 

(i)

 

\frac{2tan 30^\circ}{1+tan^2 30^\circ}

 

(A) sin 60^\circ           (B) cos 60^\circ         (C) tan 60^\circ             (D) sin 30^\circ

 

 

Solution:

 

\frac{2tan 30^\circ}{1+tan^2 30^\circ}

 

=\frac{2.\frac{1}{\sqrt{3}}}{1+(\frac{1}{\sqrt{3}})^2}=\frac{2.\frac{1}{\sqrt{3}}}{\frac{4}{3}}

 

=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{3}{2\sqrt{3}}

 

=\frac{3}{2\sqrt{3}} \times \frac{2\sqrt{3}}{2\sqrt{3}}=\frac{6\sqrt{3}}{12}=\frac{\sqrt{3}}{2}

 

sin 60^\circ is also equal to \frac{\sqrt{3}}{2}.

 

Therefore, answer is (A).


 

(ii)

 

\frac{1-tan^2 45^\circ}{1+tan^2 45^\circ}

 

(A) tan 90^\circ       (B) 1             (C) sin 45^\circ                 (D) 0

 

Solution:

 

=\frac{1-(1)^2}{1+(1)^2}=\frac{0}{1}=0

 

Therefore, the answer is (D)


 

(iii)

 

sin 2A = 2 sin A is true when A =

 

(A) 0^\circ        (B) 30^\circ         (C) 45^\circ            (D) 60^\circ

 

Solution:

 

sin 2(0) = sin 0^\circ = 0

 

2 sin (0) = 2 x 0 = 0

 

Therefore, answer is (A).


 

(iv)

 

\frac{2tan 30^\circ}{1-tan^2 30^\circ}

 

(A) cos 60^\circ      (B) sin 60^\circ            (C) tan 60^\circ             (D) sin 30^\circ

 

Solution:

 

\frac{2tan 30^\circ}{1-tan^2 30^\circ}

 

=\frac{2.\frac{1}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}})^2}=\frac{2.\frac{1}{\sqrt{3}}}{\frac{2}{3}}

 

=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\frac{3}{\sqrt{3}}=\sqrt{3}

 

tan 60^\circ is also equal to \sqrt{3}

 

Therefore, the answer is (C).


 


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