find the value of x:

Solve this identity, where the angles involved are acute angles.

Bas_tet [7]

We have proven that the trigonometric identity [(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] equals 1 + (secθ * cosec θ)

<h3>How to solve Trigonometric Identities?</h3>

We want to prove the trigonometric identity;

[(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] = 1 + sec θ

The left hand side can be expressed as;

[(tan θ)/(1 - (1/tan θ)] + [(1/tan θ)/(1 - tan θ)]

⇒ [tan²θ/(tanθ - 1)] - [1/(tan θ(tanθ - 1)]

Taking the LCM and multiplying gives;

(tan³θ - 1)/(tanθ(tanθ - 1))

This can also be expressed as;

(tan³θ - 1³)/(tanθ(tanθ - 1))

By expansion of algebra this gives;

[(tanθ - 1)(tan²θ + tanθ.1 + 1²)]/[tanθ(tanθ(tanθ - 1))]

Solving Further gives;

(sec²θ + tanθ)/tanθ

⇒ sec²θ * cotθ + 1

⇒ (1/cos²θ * cos θ/sin θ) + 1

⇒ (1/cos θ * 1/sin θ) + 1

⇒ 1 + (secθ * cosec θ)

Read more about Trigonometric Identities at; brainly.com/question/7331447

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3 0

2 years ago

Read 2 more answers

Solve the radical equation x – 7 = square root of -4x+28. Which statement is true about the solutions to the radical equation? T

Julli [10]

Answer:

x = 3 and 7

There are two true solutions.

Step-by-step explanation:

To solve $x-7 = \sqrt{-4x+28}$ , use inverse operations by squaring both sides of the equal sign.

$(x-7)^2 = (\sqrt{-4x+28})^2\\x^2-14x+49 = -4x+28\\x^2-14x+4x+49-28 = 0\\x^2-10x+21=0$

The quadratic expression can be factored into binomials and set equal to 0 by the zero product property to find x.

(x - 3) ( x - 7) = 0

x-3 = 0 so x=3

x-7 = 0 so x=7

Now check each solution into the original equation to be sure it solve the solution and is not extraneous.

$7-7 = \sqrt{-4(7)+28}\\ 0=0$

and

$3-7 = \sqrt{-4(3)+28}\\[tex]-4 = \sqrt{16}\\-4 =-4$ [/tex]

3 0

3 years ago

Write the slope-intercept form of the equation of each line

Vera_Pavlovna [14]

First let's talk about the blue line.

You can see its rising so its slope is certainly positive. But by how much is it rising? You can observe that each unit it rises it goes 1 forward and 1 up so its slope is the ratio of 1 up and 1 forward which is just 1.

We have thusly,

$y=x+n$

Now look at where blue line intercepts y-axis, -1. That is our n.

So the blue line has the equation of,

$y=x-1$

Next the black lines. The black lines are axes so their equations are a bit different.

First let's deal with x-axis, does it have slope? Yes but it is 0. The x-axis is still, not rising nor falling. Where does x-axis intercept y-axis? At 0. So the equation would be,

$y=0x+0=0$

Now we have y-axis. Does y axis have a slope? Yes but it is $\infty$ . The y-axis rises infinitely in no run. Where does it intercept y-axis? Everywhere! So what should the equation be? What if we ask where does y-axis intercept x-axis and write its equation in terms of x. Y-axis intercepts x-axis at 0 which means its equation is,