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vladimir2022 [97]

3 years ago

11

Prove that :

2 answers:

anyanavicka [17]3 years ago

7 0

Answer:

$Question 3sinθ/ (1 - cotθ) + cosθ / (1 - tanθ) =sinθ / (1 - cosθ/sinθ) + cosθ / (1 - sinθ/cosθ) =sinθ/ [(sinθ - cosθ) / sinθ] + cosθ / [(cosθ - sinθ)/cosθ] =sin²θ/ (sinθ - cosθ) + cos²θ/(cosθ - sinθ) =sin²θ/ (sinθ - cosθ) - cos²θ/(sinθ - cosθ) =(sin²θ - cos²θ) / (sinθ - cosθ) =(sinθ + cosθ)(sinθ - cosθ) / (sinθ - cosθ) =sinθ + cosθ :D$

Step-by-step explanation:

Dimas [21]3 years ago

6 0

Answer:

<h3>Question 1</h3>

sec²θ + cosec²θ =
1/cos²θ + 1/sin²θ =
(sin²θ + cos²θ)/(sin²θcos²θ) =
1 / (sin²θcos²θ) =
[(sin²θ + cos²θ)/sinθcosθ]² =
(sinθ/cosθ + cosθ/sinθ)² =
(tanθ + cotθ)²

<h3>Question 2</h3>

(1 - tan²θ) / (1 + tan²θ) =
(1 - sin²θ/cos²θ) / (1 + sin²θ/cos²θ) =
(cos²θ - sin²θ) / (cos²θ + sin²θ) =
(cosθ + sinθ)(cosθ - sinθ) / 1 =
(cosθ + sinθ)(cosθ - sinθ)

<h3>Question 3</h3>

sinθ/ (1 - cotθ) + cosθ / (1 - tanθ) =
sinθ / (1 - cosθ/sinθ) + cosθ / (1 - sinθ/cosθ) =
sinθ/ [(sinθ - cosθ) / sinθ] + cosθ / [(cosθ - sinθ)/cosθ] =
sin²θ/ (sinθ - cosθ) + cos²θ/(cosθ - sinθ) =
sin²θ/ (sinθ - cosθ) - cos²θ/(sinθ - cosθ) =
(sin²θ - cos²θ) / (sinθ - cosθ) =
(sinθ + cosθ)(sinθ - cosθ) / (sinθ - cosθ) =
sinθ + cosθ

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2 years ago

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choli [55]

Answer:

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Step-by-step explanation:

Given that;

simple interest rate r = 180% p.a

Let the amount borrowed (principal) be $x.

time t = 4 weeks = 28 days

= 28/365 year

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we all know that :

$Simple \ \ interest = \dfrac{P*r*t}{100}$

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3 years ago