Question

Solve this :-

$\displaystyle \dfrac{1 - \sin^2 \dfrac{ \pi}{6} }{1 + \sin^2 \dfrac{\pi}{4} } \times \dfrac{ \cos^2 \dfrac{\pi}{3} + \cos^2 \dfrac{\pi}{6} }{\csc^2 \dfrac{\pi}{2} - \cot^2 \dfrac{\pi}{2} } \: \div (\sin \dfrac{\pi}{3} \tan \dfrac{\pi}{6} ) + (\sec^2 \dfrac{\pi}{6} - \tan^2 \dfrac{\pi}{6} )$


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Answer 1

Answer:

2

Step-by-step explanation:

Given :

$\displaystyle \dfrac{1 - \sin^2 \dfrac{ \pi}{6} }{1 + \sin^2 \dfrac{\pi}{4} } \times \dfrac{ \cos^2 \dfrac{\pi}{3} + \cos^2 \dfrac{\pi}{6} }{\csc^2 \dfrac{\pi}{2} - \cot^2 \dfrac{\pi}{2} } \: \div (\sin \dfrac{\pi}{3} \tan \dfrac{\pi}{6} ) + (\sec^2 \dfrac{\pi}{6} - \tan^2 \dfrac{\pi}{6} )$

$\pi = 180^{\circ}$

So we know that pi is 180°,

$\frac{\pi}{6} = 30^{^{\circ}}$

$\frac{\pi}{4} = 45^{^{\circ}}$

$\frac{\pi}{3} = 60^{^{\circ}}$

$\frac{\pi}{2} = 90^{^{\circ}}$

So rewriting the above terms:

$\displaystyle \dfrac{1 - \sin^2 30 }{1 + \sin^2 45 } \times \dfrac{ \cos^2 60+ \cos^2 30 }{\csc^2 90 - \cot^2 90} \: \div (\sin 60 \tan 30) + (\sec^2 30 - \tan^2 30)$

Now Substituting the values of the trigonometric values would yield us the answer

$\sin(30) = \frac{1}{2}$

$\sin(45) = \frac{1}{ \sqrt{2} }$

$\cos(60) = \frac{1}{2}$

$\cos(30) = \frac{ \sqrt{3} }{2}$

$\csc(90) = 1$

$\cot(90) = 0$

$\sin(60) = \frac{ \sqrt{3} }{2}$

$\tan(30) = \frac{1}{ \sqrt{3} }$

$\sec(30) = \frac{2}{ \sqrt{3} }$

$\tan(30) = \frac{1}{ \sqrt{3} }$

Now Substituting the values :

$\displaystyle \dfrac{1 - ( \frac{1}{2} ) ^{2} }{1 + ( \frac{1}{ \sqrt{2} })^{2} } \times \dfrac{ \ (\frac{1}{2})^{2} + (\frac{ \sqrt{3} }{2})^{2} }{ {1}^{2} - \ {0}^{2} } \: \div ( \frac{ \sqrt{3} }{2} \times \frac{1}{ \sqrt{3} } ) + ( (\frac{2}{ \sqrt{3} } ^{2} )- ( \frac{1}{ \sqrt{3} } ) ^{2} )$

$\dfrac{1}{2} \times 1 \: \div \dfrac{1}{2} + 1$

$\dfrac{1}{2} \times \dfrac{2}{1} + 1$

= 1+1

=2

The value is equal to 2

Answer 2

Answer

$\rm \frac{1-sin^2\frac{\pi}{6}}{1+sin^2\frac{\pi}{4}}\times \frac{cos^2\frac{\pi}{3}+cos^2\frac{\pi}{6}}{csc^2\frac{\pi}{2}-cot^2\frac{\pi}{2}} \div \left( sin\frac{\pi}{3}tan\frac{\pi}{6} \right)+\left( sec^2\frac{\pi}{6}-tan^2\frac{\pi}{6} \right)\\\\:\to \rm \frac{1-sin^2(30)}{1+sin^2(45)}\times \frac{cos^2(60)+cos^2(30)}{csc^2(90)-cot^2(90)}\div (sin(60).tan(30)+(sec^2(30)-tan^2(30))$

$:\to \rm \frac{1-\frac{1}{4}}{1+\frac{1}{2}}\times \frac{\frac{1}{4}+\frac{3}{4}}{1-0}\div (\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{3}})+(\frac{4}{3}-\frac{1}{3})\\\\:\to \rm \frac{\frac{3}{4}}{\frac{3}{2}}\times \frac{1}{1}\div (\frac{1}{2})+(1)\\\\:\to \rm \frac{1}{2}\times (2)+1\\\\:\to \rm 1+1\\\\:\to \rm 2$

Danger points

→ Be aware of Bodmas rule

where ,

B denotes Brackets

o denotes of

d denotes division

m denotes multiplication

a denotes addition

s denotes subtraction