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Misha Larkins [42]

2 years ago

14

1+sinx+cosx/1+sinx-cosx=1+cosx/sinx help pls by proving

1 answer:

Monica [59]2 years ago

6 0

Answer:

Below.

Step-by-step explanation:

I'll write sin x as s and cos x as c so we have:

(1 + s +c)/(1 + s - c) = (1 + c)/s

Cross multiplying:

s + s^2 + cs = 1 + s - c + c + cs - c^2

s + s^2 + cs = 1 + s + cs - c^2

s^2 + c^2 + s - s + cs - cs = 1

s^2 + c^2 = 1.

- that is sin^2 x + cos^2 x = 1 which is a known identity.

Therefore the original identity is proved.

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Identify the solution of the inequality |4x − 2| > 14 and the graph that represents it.

kodGreya [7K]

The solution to the inequality |4x − 2| > 14 is x > 4 and x < -3

<h3>What is an inequality?</h3>

An inequality is an expression that shows the non equal comparison of two or more numbers and variables.

Given the inequality:

|4x − 2| > 14

Hence:

4x - 2 > 14 or -(4x - 2) > 14

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The solution to the inequality |4x − 2| > 14 is x > 4 and x < -3

Find out more on inequality at: brainly.com/question/24372553

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1 year ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

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