Question

can someone please help me with this trigonometry problem!​

Answer 1

cosec A = 2

cosec A = cosec 30°

A = 30°

$\frac{1}{tan \: a} + \frac{sin \: a}{1 + cos \: a} \\ \\ = \: \frac{1}{tan \: 30} + \frac{sin \: 30}{1 + cos \: 30} \\ \\ = \frac{1}{ \frac{1}{ \sqrt{3} } } + \frac{ \frac{1}{2} }{1 + \frac{ \sqrt{3} }{2} } \\ \\ = \sqrt{3} + \frac{ \frac{1}{2} }{ \frac{2 + \sqrt{3} }{2} } \\ \\ = \sqrt{3} + \frac{1}{2 + \sqrt{3} } \\ \\ = \frac{ \sqrt{3} (2 + \sqrt{3}) + 1 }{2 + \sqrt{3} } \\ \\ = \frac{2 \sqrt{3} + 3 + 1 }{2 + \sqrt{3} } \\ \\ = \frac{2 \sqrt{3} + 4 }{2 + \sqrt{3} } \\ \\ = \frac{2 \sqrt{3} + 4 }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ = \frac{(2 \sqrt{3} + 4)(2 - \sqrt{3}) }{ {(2)}^{2} - {( \sqrt{3}) }^{2} } \\ \\ = \frac{2 \sqrt{3} (2 - \sqrt{3}) + 4(2 - \sqrt{3} )}{4 - 3} \\ \\ = \frac{4 \sqrt{3} - 6 + 8 - 4 \sqrt{3} }{1} \\ \\ = 2$

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Answer 2

2

Step-by-step explanation:

Given:

cosec A = 2

To find:

The value of

$\dfrac{1}{ \tan A} + \dfrac{ \sin A}{1 + \cos A}$

Solution:

cosec A = 2

As, We know

cosec 30° = 2

So, cosec A = cosec 30°

A = 30°

Now,

$\frac{1}{ \tan A} + \frac{ \sin A}{1 + \cos A} \\ \\ \frac{1}{ \tan30} + \frac{ \sin30}{1 + \cos 30} \\ \\ \cot30 + \frac{ \frac{1}{2} }{1 + \frac{ \sqrt{3} }{2} } \\ \\ \sqrt{3} + \frac{ \frac{1}{2} }{ \frac{2 + \sqrt{3} }{2} } \\ \\ \sqrt{3} + \frac{1}{2} \times \frac{2}{2 + \sqrt{3} } \\ \\ \sqrt{3} + \frac{1}{2 + \sqrt{3} } \\ \\ [\text{Rationalizing the denominator}] \\ \\ \sqrt{3} \: + \frac{(2 - \sqrt{3}) }{(2 + \sqrt{3} )(2 - \sqrt{3} )} \\ \\ \sqrt{3} + \frac{2 - \sqrt{3} }{(2){}^{2} - {( \sqrt{3} )}^{2} } \\ \\ \sqrt{3} + \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \sqrt{3} + 2 - \sqrt{3} \\ \\ 2$

Hence, The required answer is 2.

Some Important trigonometry Rules:

sinø . cosecø = 1

cosø . secø = 1

tanø . cotø = 1

sin²ø + cos²ø = 1

cosec²ø - cot² = 1

sec²ø - tan²ø = 1

Trigonometry Value

See in the attachment.