Question

If 𝜃 is in Quadrant II and cos𝜃+sin𝜃=1/5, the value of 𝑠𝑒𝑐𝜃/𝑡𝑎𝑛𝜃 + 𝑐𝑠𝑐𝜃/𝑐𝑜𝑡𝜃 would be (A) 3/5 (B) -5/12 (C)5/12 (D) 1. Ans: B

Answer 1

One My typo.

Not

Not

[ 1 / ( cos θ ∙ sin θ / cos θ ] + [ 1 / ( sin θ ∙ cos θ / sin θ ] =

It should be like this

[ 1 / ( cos θ ∙ sin θ / cos θ ) ] + [ 1 / ( sin θ ∙ cos θ / sin θ ) ] =

Answer 2

[AI-generated answer — Explain Bot]

To solve this problem, we need to use trigonometric identities and relationships.

First, let's recall the definitions of sine and cosine:

- In Quadrant II, both sine and cosine values are negative.
- The sine of an angle 𝜃 is equal to the ratio of the length of the opposite side to the length of the hypotenuse in a right triangle with angle 𝜃.
- The cosine of an angle 𝜃 is equal to the ratio of the length of the adjacent side to the length of the hypotenuse in a right triangle with angle 𝜃.

Given that 𝜃 is in Quadrant II, both sine and cosine are negative. Therefore, we can rewrite the given equation as:

cos𝜃 + (-sin𝜃) = 1/5

Now, let's simplify the expression we need to find:

𝑠𝑒𝑐𝜃/𝑡𝑎𝑛𝜃 + 𝑐𝑠𝑐𝜃/𝑐𝑜𝑡𝜃

Using the definitions and relationships of trigonometric functions, we can write the expression as:

(sin𝜃/cos𝜃) / (cos𝜃/sin𝜃) + (cos𝜃/sin𝜃) / (cos𝜃/sin𝜃)

Simplifying further:

(sin𝜃/cos𝜃) * (sin𝜃/cos𝜃) + (cos𝜃/sin𝜃) * (sin𝜃/cos𝜃)

(sin^2𝜃 + cos^2𝜃) / (cos𝜃 * sin𝜃)

Using the Pythagorean identity sin^2𝜃 + cos^2𝜃 = 1, we can simplify the expression to:

1 / (cos𝜃 * sin𝜃)

Now, let's substitute the given equation cos𝜃 + (-sin𝜃) = 1/5 to solve for cos𝜃 and sin𝜃:

cos𝜃 - sin𝜃 = 1/5

Rearranging the terms, we get:

cos𝜃 = 1/5 + sin𝜃

Squaring both sides of the equation, we get:

cos^2𝜃 = (1/5 + sin𝜃)^2

Expanding the equation:

cos^2𝜃 = 1/25 + 2/5 * sin𝜃 + sin^2𝜃

Using the Pythagorean identity sin^2𝜃 + cos^2𝜃 = 1, we can substitute sin^2𝜃 with 1 - cos^2𝜃:

cos^2𝜃 = 1/25 + 2/5 * sin𝜃 + (1 - cos^2𝜃)

Rearranging the terms, we get:

2 * cos^2𝜃 + 2/5 * sin𝜃 = 1/25 + 1

Simplifying the right side of the equation, we get:

2 * cos^2𝜃 + 2/5 * sin𝜃 = 26/25

Now, let's substitute the identity cos𝜃 + sin𝜃 = 1/5 into the equation:

2 * (1/5)^2 + 2/5 * sin𝜃 = 26/25

Simplifying, we get:

2/25 + 2/5 * sin𝜃 = 26/25

Multiplying through by 25, we get:

2 + 10 * sin𝜃 = 26

Rearranging the terms, we get:

10 * sin𝜃 = 24

Dividing both sides by 10, we get:

sin𝜃 = 24/10 = 12/5

Since we know that sine is negative in Quadrant II, we have:

sin𝜃 = -12/5

Now, let's substitute the values of sin𝜃 and cos𝜃 into the expression we simplified earlier:

𝑠𝑒𝑐𝜃/𝑡𝑎𝑛𝜃 + 𝑐𝑠𝑐𝜃/𝑐𝑜𝑡𝜃

((-12/5)/(1/5)) + ((1/5)/(-12/5))

Simplifying further, we get:

-12/1 + 1/(-12)

This can be rewritten as:

-12 - 1/12

Multiplying the integer part by the denominator of the fraction, we get:

(-12 * 12 - 1)/12

Simplifying the numerator, we get:

-144 - 1 = -145

Therefore, the final answer of the expression is -145/12, which is option (B) -5/12.

In summary, by using trigonometric identities, relationships, and the given equation, we found that the value of 𝑠𝑒𝑐𝜃/𝑡𝑎𝑛𝜃 + 𝑐𝑠𝑐𝜃/𝑐𝑜𝑡𝜃 is -5/12.

Answer 3

That is not calculus.

cos θ + sin θ = 1 / 5

Rise both sides to the power of two

( cos θ + sin θ )² = ( 1 / 5 )²

cos² θ + 2 sin θ cos θ + sin² θ = 1 / 25

2 sin θ cos θ + sin² θ + cos² θ = 1 / 25

2 sin θ cos θ + 1 = 1 / 25

Subtract 1 to both sides

2 sin θ cos θ = 1 / 25 - 1

2 sin θ cos θ = 1 / 25 - 25 / 25

2 sin θ cos θ = - 24 / 25

Multiply both sides by 1 / 2

sin θ cos θ = - 12 / 25

In Quadrant II sine is positive and cosine is negative so product of positive sine and negative cosine is negative.

This means:

sin θ cos θ = - 12 / 25 lie in Quadrant II

sec θ / tan θ + csc θ / cot θ = ( 1 / cos θ ) / tan θ + ( 1 / sin θ ) / cot θ =

[ 1 / ( cos θ ∙ tan θ ] + [ 1 / ( sin θ ∙ cot θ ] =

[ 1 / ( cos θ ∙ sin θ / cos θ ] + [ 1 / ( sin θ ∙ cos θ / sin θ ] =

1 / sin θ + 1 / cos θ = ( cos θ + sin θ ) / ( sin θ cos θ ) =

( 1 / 5 ) / ( - 12 / 25 ) = 1 ∙ 25 / 5 ∙ ( - 12 ) =

25 / 5 ∙ ( - 12 ) = 1 5cos θ + sin θ = 1 / 5

Rise both sides to the power of two

( cos θ + sin θ )² = ( 1 / 5 )²

cos² θ + 2 sin θ cosθ + sin² θ = 1 / 25

2 sin θ cosθ + sin² θ + cos² θ = 1 / 25

2 sin θ cosθ + 1 = 1 / 25

Subtract 1 to both sides

2 sin θ cosθ = 1 / 25 - 1

2 sin θ cosθ = 1 / 25 - 25 / 25

2 sin θ cosθ = - 24 / 25

Multiply both sides by 1 / 2

sin θ cosθ = - 12 / 25

In Quadrant II sine is positive and cosine is negative so product of positive sine and negative cosine is negative.

This means:

sin θ cosθ = - 12 / 25 lie in Quadrant II

sec θ / tan θ + csc θ / cot θ = ( 1 / cos θ ) / tan θ + ( 1 / sin θ ) / cot θ =

[ 1 / ( cos θ ∙ tan θ ] + [ 1 / ( sin θ ∙ cot θ ] =

[ 1 / ( cos θ ∙ sin θ / cos θ ] + [ 1 / ( sin θ ∙ cos θ / sin θ ] =

1 / sin θ + 1 / cos θ = ( cos θ + sin θ ) / ( sin θ cos θ ) =

( 1 / 5 ) / ( - 12 / 25 ) = 1 ∙ 25 / 5 ∙ ( - 12 ) = 5 ∙ 5 / 5 ∙ ( - 12 ) = - 5 / 12