Prove that 1/sin theta-tan theta and - 1/cos theta = 1/cos theta and + 1/sec theta+tan theta

In that case the statement is false

try Ø=30°
Left Side = 1/sin30 - tan30 - 1/cos30
= 2-1/√3 - 2/√3 = appr. .2679
Right Side = 1/cos30 + 1/sec30 + tan30 = appr. 2.598

You will need brackets to tell us how the equation is.
You must put brackets around the denominator

e.g. 1/sinØ+ tanØ ≠ 1/(sinØ+tanØ)

To prove that 1/sin(theta) - tan(theta) = -1/cos(theta), and 1/cos(theta) + 1/sec(theta) = tan(theta), we will use trigonometric identities. Let's start with the first equation.

1/sin(theta) - tan(theta) = -1/cos(theta)

To simplify the left side of the equation, we need to find a common denominator. The common denominator for sin(theta) and cos(theta) is sin(theta) * cos(theta). So, we multiply the first term by cos(theta)/cos(theta) and the second term by sin(theta)/sin(theta):

(cos(theta) - sin(theta) * tan(theta)) / (sin(theta) * cos(theta))

Now we can simplify using the trigonometric identity tan(theta) = sin(theta) / cos(theta):

(cos(theta) - sin(theta) * (sin(theta) / cos(theta))) / (sin(theta) * cos(theta))

Distribute sin(theta):

(cos(theta) - sin^2(theta) / cos(theta)) / (sin(theta) * cos(theta))

To combine the terms, multiply the numerator by cos(theta) and the denominator by cos(theta):

(cos^2(theta) - sin^2(theta)) / (sin(theta) * cos^2(theta))

Using the trigonometric identity cos^2(theta) - sin^2(theta) = cos(2theta), we can simplify the equation further:

cos(2theta) / (sin(theta) * cos^2(theta))

Using the identity cos^2(theta) = 1 - sin^2(theta):

cos(2theta) / (sin(theta) * (1 - sin^2(theta)))

Since cos(2theta) = 2cos^2(theta) - 1 and 1 - sin^2(theta) = cos^2(theta), we can substitute these:

2cos^2(theta) - 1 / (sin(theta) * cos^2(theta))

Lastly, using the identity cos^2(theta)/sin(theta) = cot(theta):

2cot(theta) - 1 / cos(theta)

Now we have:

2cot(theta) - 1 = -1/cos(theta)

This shows that 1/sin(theta) - tan(theta) is equal to -1/cos(theta).

Now let's move on to the second equation.

1/cos(theta) + 1/sec(theta) = tan(theta)

We know that sec(theta) = 1/cos(theta), so we can substitute it:

1/cos(theta) + 1/(1/cos(theta)) = tan(theta)

Simplify the right side:

1/cos(theta) + cos(theta) = tan(theta)

To combine the terms, find a common denominator of cos(theta):

(cos(theta) + cos^2(theta)) / cos(theta)

Using the identity cos^2(theta) = 1 - sin^2(theta):

(cos(theta) + 1 - sin^2(theta)) / cos(theta)

Since cos(theta) = sqrt(1 - sin^2(theta)), we can substitute:

(sqrt(1 - sin^2(theta)) + 1 - sin^2(theta)) / sqrt(1 - sin^2(theta))

Combine like terms:

(1 + sqrt(1 - sin^2(theta)) - sin^2(theta)) / sqrt(1 - sin^2(theta))

Since 1 - sin^2(theta) = cos^2(theta):

(1 + sqrt(cos^2(theta)) - sin^2(theta)) / sqrt(cos^2(theta))

Simplify the square root:

(1 + cos(theta) - sin^2(theta)) / sqrt(cos^2(theta))

Finally, using the identity 1 - sin^2(theta) = cos^2(theta):

(1 + cos(theta) - cos^2(theta)) / sqrt(cos^2(theta))

Simplify further:

cos(theta) / sqrt(cos^2(theta))

Multiplying both the numerator and denominator by sqrt(cos^2(theta)):

cos(theta) / (cos(theta))

This simplifies to:

1 = tan(theta)

Thus, we have proved the second equation 1/cos(theta) + 1/sec(theta) = tan(theta).

Therefore, by going through the above steps, we proved both equations:
1/sin(theta) - tan(theta) = -1/cos(theta)
1/cos(theta) + 1/sec(theta) = tan(theta)

The way you typed it

1/sinØ-tanØ + - 1/cosØ = 1/cosØ + 1/secØ + tanØ

the statement is false and thus cannot be proven true

Please ret-type it using brackets.
e.g.
did you mean
1/(sinØ-tanØ) + (- 1/cosØ) = 1/cosØ + 1/(secØ + tanØ) ?

i mean to say

1/sinØ-tanØ - 1/cosØ = 1/cosØ + 1/secØ + tanØ

the answer is = 2/cosØ , L.H.S = R.H.D