Question

Evaluate the following
1) sin 45't cos45

2) cos45/ sec30+ cosec 60


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Answer 1

Answer:

cos(90-45)/sec(90-30)+cosec(90-60)

cos(45)/ sec(60)+cosec(30)

cos(45)/ sec(2)

45/2

22.5

hope it is useful to u

PLZZ take as BRAIN least

Answer 2

Answer:

(i) sin 45° + cos 45°

= $\frac{1}{\sqrt{2} } +\frac{1}{\sqrt{2} }$ [∵ sin 45° and cos 45° = $\frac{1}{\sqrt{2} }$]

=$\frac{1+1}{\sqrt{2} }$

= $\frac{2}{\sqrt{2} }$

(ii)  $\frac{cos 45^{o} }{sec 30^{o} csc 60^{o} }$

⇒ We know that,

= cos 45° = $\frac{1}{\sqrt{2} }$

= sec 30° = $\frac{2}{\sqrt{3} }$

= csc 60° = $\frac{2}{\sqrt{3} }$

= $\frac{\frac{1}{\sqrt{2} } }{\frac{2}{\sqrt{3} } + \frac{2}{\sqrt{3} } } = \frac{\frac{1}{\sqrt{2} } }{\frac{2+2}{\sqrt{3} } }$

= $\frac{\frac{1}{\sqrt{2} } }{\frac{4}{\sqrt{3} } }$ = 1 x √3/4 x √2 (FOR YOUR UNDERSTANDING ⇒ 'x' refers to multiplication and you should write 1,√3 divided by 4,√2)

⇒ $\frac{\sqrt{3} }{4\sqrt{2} }$