Question

Class 10th trigonometry​

Answer 1

Solution

$\sf \dfrac{cos \: 58 \degree}{sin \: 32\degree} + \dfrac{sin \: 22 \degree}{cos \: 68\degree} - \dfrac{cos \: 38 \degree \: cosec\: 52 \degree}{tan \: 18\degree \: tan\: 35 \degree \: tan \:60 \degree \: tan\: 72 \degree \: tan\: 55 \degree}$

$\sf = \dfrac{cos \: 58 \degree}{sin \: (90 \degree - 58\degree)} + \dfrac{sin \: 22 \degree}{cos \: (90 \degree - 22\degree)} - \dfrac{cos \: 38 \degree \: cosec\: (90 \degree - 38 \degree)}{(tan \: 18\degree \: tan\: 72 \degree) \:( tan \:35 \degree \: tan\: 55 \degree) \: tan\: 60 \degree}$

we know that,

◉ sin θ = cos(90 - θ)

◉ tan θ = cot(90 - θ)

◉ sec θ = cosec(90 - θ)

Using this equations we get,

$\sf = \dfrac{cos \: 58 \degree}{cos \: 58\degree} + \dfrac{sin \: 22 \degree}{sin \: 22\degree} - \dfrac{cos \: 38 \degree \: sec\: 38\degree}{ \{tan \: 18\degree \: (tan\:90 \degree - 18 \degree) \} \{ tan \:35\degree \: (tan\: 90 \degree - 35 \degree) \} \: tan\: 60\degree}$

$\sf = 1 + 1 - \dfrac{1}{(tan \: 18 \degree \: cot \: 18 \degree)(tan \: 35 \degree \: cot \: 35 \degree)tan \: 60 \degree}$

$\sf = 2 - \dfrac{1}{(1)(1) \sqrt{3} }$

$\sf = 2 - \dfrac{1}{ \sqrt{3} }$

$\sf = \dfrac{2 \sqrt{3} - 1 }{ \sqrt{3} }$

$\sf = \dfrac{2 \sqrt{3} - 1 }{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\sf = \dfrac{(6 - \sqrt{3} ) }{3}$

Answer 2

Step-by-step explanation:

Solution

\sf \dfrac{cos \: 58 \degree}{sin \: 32\degree} + \dfrac{sin \: 22 \degree}{cos \: 68\degree} - \dfrac{cos \: 38 \degree \: cosec\: 52 \degree}{tan \: 18\degree \: tan\: 35 \degree \: tan \:60 \degree \: tan\: 72 \degree \: tan\: 55 \degree}

sin32°

cos58°

+

cos68°

sin22°

−

tan18°tan35°tan60°tan72°tan55°

cos38°cosec52°

\sf = \dfrac{cos \: 58 \degree}{sin \: (90 \degree - 58\degree)} + \dfrac{sin \: 22 \degree}{cos \: (90 \degree - 22\degree)} - \dfrac{cos \: 38 \degree \: cosec\: (90 \degree - 38 \degree)}{(tan \: 18\degree \: tan\: 72 \degree) \:( tan \:35 \degree \: tan\: 55 \degree) \: tan\: 60 \degree}=

sin(90°−58°)

cos58°

+

cos(90°−22°)

sin22°

−

(tan18°tan72°)(tan35°tan55°)tan60°

cos38°cosec(90°−38°)

we know that,

◉ sin θ = cos(90 - θ)

◉ tan θ = cot(90 - θ)

◉ sec θ = cosec(90 - θ)

Using this equations we get,

\sf = \dfrac{cos \: 58 \degree}{cos \: 58\degree} + \dfrac{sin \: 22 \degree}{sin \: 22\degree} - \dfrac{cos \: 38 \degree \: sec\: 38\degree}{ \{tan \: 18\degree \: (tan\:90 \degree - 18 \degree) \} \{ tan \:35\degree \: (tan\: 90 \degree - 35 \degree) \} \: tan\: 60\degree}=

cos58°

cos58°

+

sin22°

sin22°

−

{tan18°(tan90°−18°)}{tan35°(tan90°−35°)}tan60°

cos38°sec38°

\sf = 1 + 1 - \dfrac{1}{(tan \: 18 \degree \: cot \: 18 \degree)(tan \: 35 \degree \: cot \: 35 \degree)tan \: 60 \degree}=1+1−

(tan18°cot18°)(tan35°cot35°)tan60°

1

\sf = 2 - \dfrac{1}{(1)(1) \sqrt{3} }=2−

(1)(1)

3

1

\sf = 2 - \dfrac{1}{ \sqrt{3} }=2−

3

1

\sf = \dfrac{2 \sqrt{3} - 1 }{ \sqrt{3} }=

3

2

3

−1

\sf = \dfrac{2 \sqrt{3} - 1 }{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }=

3

2

3

−1

×

3

3

\sf = \dfrac{(6 - \sqrt{3} ) }{3}=

3

(6−

3

)