Math, asked by kanika2749, 10 months ago

find the principal and general solution of cosec x = -2​

Answers

Answered by RvChaudharY50
109

Question :-- find the principal and general solution of cosec x = -2 ?

Concept used :--

→ -1≤y≤ 1, that is, the value of y is between -1 and 1, so there is a solution.

The set of all solutions to sin(x) = y is

x = sin^(-1)y + 2kπ and x = −sin^(-1)y + (2k+1)π,

where k can be any integer; that is, the solutions for x consist of sin^(-1)y plus all even multiples of π, together with minus sin-1(y) plus all odd multiples of π.

→ The principal value of sin^(−1) x is α, where - π/2 ≤ α ≤ π/2 and α satisfies the equation sin θ = x.

→ cosec x = 1/sinx

→ sin30° = 1/2

→ sin150° = -1/2

→ sin(π+x) = - sinx

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Solution :----

Given , cosec x = (-2)

→ 1/sinx = (-2)

Cross - Multiplying we get,

→ sinx = (-1/2)

→ sinx = - sin30°

→ sinx = - sin(π/6)

→ sinx = sin(π+π/6)

→ sinx = sin(7π/6)

Therefore in general, sin^(−1)(-1/2) = θ = nπ + (-1)^n π/6 and the angle nπ + (- 1)^n 7π/6 is called the general value of sin^(-1)(-1/2).

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Now , The positive or negative least numerical value of the angle is called the principal value ...

So,

==>> Principal value of sinx = sin(7π/6) = - π/2 ≤ (-π/6) ≤ π/2

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Answered by nehu215
1

Step-by-step explanation:

Using this logic ,In right ΔABC

sine of ∠ A = P/H = BC/AC

cosine of ∠ A = B/H = AB/AC

tangent of ∠ A = P/B = BC/AB

cosecant of ∠ A = 1/ sine of A = AC/BC

secant of ∠ A = 1/cosine of A = AC/AB

cotangent of ∠ A = 1/tangent of A = AB/BC

The value of sin A or cos A never exceeds 1, whereas the value of sec A or cosec A is always greater than or equal to 1.

Trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.

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