Question

Maths experts please solve:-

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Answer 1

(i)

$\frac{sin \: {70}^{0} }{cos \: {20}^{0} } + \frac{cosec \: {36}^{0} }{sec \: {54}^{0} } - \frac{2cos \: {43}^{0} \: cosec \: {47}^{0} }{tan \: {10}^{0} \: tan \: {40}^{0} \: tan \: {50}^{0} \: tan \: {80}^{0} }$

Using the formula,

sin (90° - θ) = cos θ

tan (90° - θ) = cot θ

sec (90° - θ) = cosec θ

Now,

$\frac{sin \: ( {90}^{0} - {20}^{0})}{cos \: {20}^{0} } + \frac{cosec \: ( {90}^{0} - {54}^{0} )}{sec \: {54}^{0} } - \frac{2cos \: {43}^{0} cosec \: ( {90}^{0} - {43}^{0} ) }{tan \: ( {90}^{0} - {80}^{0} ) \: tan \: {80}^{0} \: tan\:({90}^{0} - {40}^{0})\: tan\: {40}^{0}} \\ \frac{cos {20}^{0} }{cos \: {20}^{0} } + \frac{sec \: {54}^{0} }{sec \: {54}^{0} } - \frac{2cos \: {43}^{0} \: sec \: {43}^{0} }{cot\: {80}^{0}\: tan\: {80}^{0}\:cot\:{40}^{0}\:tan{40}^{0}}$

After solving,

1 + 1 - 2 = 0

(ii)

cos 15° = cos (45° - 30°)

= cos 45° cos 30° + sin 45° sin 30°

= 1/√2*√3/2 + 1/√2*1/2

= √3/2√2 + 1/2√2

= (√3 + 1) / 2√2

sin 15° = sin (45° - 30°)

= sin 45° cos 30° - cos 45° sin 30°

= 1/√2*√3/2 - 1/√2*1/2

= √3/2√2 - 1/2√2

= (√3 - 1) / 2√2

Hope It Helps :)