Question

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Answer 1

AnsWer :

5.

QuestioN :

The exact value of expression

$\tt tan \dfrac{ \pi}{20} - tan \dfrac{3 \pi}{20} + tan \dfrac{5 \pi}{20} - tan \dfrac{7 \pi}{20} + tan \dfrac{9 \pi}{20} \: is$

SolutioN :

$\tt : \implies tan \dfrac{ \cancel{180}}{ \cancel{20}} - tan \dfrac{3 \times \cancel{180}}{ \cancel{20}} + tan \dfrac{5 \times \cancel{180}}{ \cancel{20}} - tan \dfrac{7 \times \cancel{180}}{ \cancel{20}} + tan \dfrac{9 \times \cancel{180}}{ \cancel{20}}$

$\tt : \implies tan \: 9 - tan \: 27 + tan \: 45 - tan \: 63 + tan \: 81.$

$\tt : \implies tan \: 9 - tan \: 27 + tan \: 45 - tan \: (90 - 27) + tan \: (90 - 9).$

Apply formula.

• tan ( 90 - theta ) = cot theta.
• cot ( 90 - theta ) = tan theta.

$\tt : \implies tan \: 9 - tan \: 27 + tan \: 45 - cot \: 27 + cot \: 9.$

★ Taking One sides tan 9° , cot 9° and tan 27° , cot 27°.

$\tt : \implies (tan \: 9 + cot \: 9)- (tan \: 27 +cot \: 27) + tan \: 45.$

★ Now, tan theta = Sin theta / cos theta.

$\tt : \implies \Big( \frac{sin \: 9}{cos \: 9} + \frac{cos \: 9}{sin \: 9} \Big)- \Big( \frac{sin \: 27}{cos \: 27} + \frac{cos \: 27}{sin \: 27}\Big) + tan \: 45.$

★ tan 45° = 1.

$\tt : \implies \Big( \frac{{sin }^{2} 9 + {cos}^{2}9 }{sin9 \times cos 9} \Big)- \Big( \frac{{sin }^{2}27+ {cos}^{2}27 }{sin27 \times cos 27}\Big) + 1.$

$\tt : \implies \Big( \frac{1 }{sin9 \times cos 9} \Big)- \Big( \frac{1 }{sin27 \times cos 27}\Big) + 1.$

★ Let Multiple by 2.

$\tt : \implies \Big( \frac{1 }{ \frac{2\Big(sin9 \times cos 9\Big)}{2}}\Big)- \Big( \frac{1 }{ \frac{ 2\Big(sin27 \times cos 27\Big)}{2}}\Big) + 1.$

$\tt : \implies \Big( \frac{1 }{ \frac{\Big(sin18 \times cos 18\Big)}{2}}\Big)- \Big( \frac{1 }{ \frac{ \Big(sin54 \times cos 54\Big)}{2}}\Big) + 1.$

# Note :

• Sin 18° = √5 - 1 /4.
• Sin 54° = √5 + 1/4.

$\tt : \implies \dfrac{2}{ \frac{ \sqrt{5} - 1 }{4} } - \dfrac{2}{ \frac{ \sqrt{5} + 1 }{4} } + 1.$

$\tt : \implies \dfrac{8}{ \sqrt{5} - 1} - \dfrac{8}{ \sqrt{5} + 1 } + 1$

$\tt : \implies \dfrac{8 (\sqrt{5} + 1) - 8( \sqrt{5} - 1) }{ (\sqrt{5} - 1)( \sqrt{5} + 1) } + 1$

°•° ( a - b )( a + b ) = a² - b² .

$\tt : \implies \dfrac{8 \sqrt{5} + 8- 8\sqrt{5} + 8}{ 5 - 1 } + 1$

$\tt : \implies \dfrac{\cancel{16}}{\cancel4} + 1$

$\tt : \implies 4 + 1.$

$\tt : \implies 5.$