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Trigonometric Ratios CBSE NCERT Solutions Chapter 8 Exercise 8.2 Question 1

 

Trigonometric Ratios CBSE NCERT Solutions Chapter 8 Exercise 8.2 Question 1


 

1.  Evaluate the following:

 

(i) sin 60^\circ cos 30^\circ + sin 30^\circ cos 60^\circ

 

(ii) 2 tan^2 45^\circ + cos^2 30^\circ- sin^260^\circ

 

(iii) cos 45^\circ/ (sec 30^\circ + cosec 30^\circ)

 

(iv) (sin 30^\circ + tan 45^\circ- cosec 60^\circ) / ( sec 30^\circ + cos 60^\circ + cot 45^\circ)

 

(v) (5 cos^2 60^\circ +4sec^2 30^\circ-tan^2 45^)/ (sin^2 30^\circ + cos^2 30^\circ)

 

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.


 

Solution (i)

 

sin 60^\circ cos 30^\circ + sin 30^\circ cos 60^\circ

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


 

Solution (ii)

 

2 tan^2 45^\circ + cos^2 30^\circ- sin^260^\circ

 

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

 

= 2+\frac{3}{4}-\frac{3}{4} = 2


 

Solution (iii)

 

cos 45^\circ/ (sec 30^\circ + cosec 30^\circ)

 

=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}

 

=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}}=\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2+2\sqrt{3}}

 

=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}

 

=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}} \times \frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}

 

=\frac{2\sqrt{6}-2\sqrt{18}}{8-24}=\frac{2(\sqrt{6}-\sqrt{18})}{-16}=\frac{\sqrt{6}-\sqrt{18}}{-8}

 

=\frac{\sqrt{18}-\sqrt{6}}{8}=\frac{3\sqrt{2}-\sqrt{6}}{8}


 

Solution (iv)

 

(sin 30^\circ + tan 45^\circ- cosec 60^\circ) / ( sec 30^\circ + cos 60^\circ + cot 45^\circ)

 

=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}

 

=\frac{\frac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\frac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}

 

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

 

=\frac{3\sqrt{3}-4}{4+3\sqrt{3}}

 

=\frac{3\sqrt{3}-4}{4+3\sqrt{3}} \times \frac{3\sqrt{3}-4}{3\sqrt{3}-4}

 

=\frac{27+16-24\sqrt{3}}{27-16}=\frac{43-24\sqrt{3}}{11}


 

Solution (v)

 

(5 cos^2 60^\circ +4sec^2 30^\circ-tan^2 45^)/ (sin^2 30^\circ + cos^2 30^\circ)

 

=\frac{5(\frac{1}{2})^2+4(\frac{2}{\sqrt{3}})^2-1}{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}

 

=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}

 

=\frac{\frac{15+64-12}{12}}{1}=\frac{67}{12}


 


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