Question

Is cosec theta/ cot theta=cos theta

Answer 1

No

cosec theta / cot theta = cos theta

Taking LHS

cosec theta / cot theta

1/sin theta/ cos theta / sin theta

[ cosec theta = 1/sin theta]

[ cot theta = cos theta / sin theta]

1 / sin theta ÷ cos theta / sin theta

1/ sin theta × sin theta / cos theta

1 / cos theta , which is not equal to RHS

please by use oblique sign as fraction sign then you'll get this better

All the best

May this help u

Answer 2

Answer:

What is cosec theta/cot theta + tan theta =cos theta?

This is an equation that may be true for some, or even all, values of theta. As your question has been flagged as a ‘proof’, this implies that you suspect or have been told that the equation is true for all values of θ and want us to show this.

First, the mathematical convention is that multiplications and divisions have priority over additions and subtractions, e.g. 4/1 + 3 = (4/1) + 3 = 4 + 3 = 7. In my day, this convention was taught to children before their age got into double figures. While many people remember this convention, they forget that it also applies to calculations using functions.

When using functions, I like to use parentheses, brackets or, sometimes, even braces around the argument as it often helps clarify the meaning. So, I’ld write your equation as:

cosec(θ)cot(θ)+tan(θ)=cos(θ)

As cot(θ) is the multiplicative inverse of tan(θ) , dividing by cot(θ) is the same as multiplying by tan(θ) , thus we can rewrite the left side of your equation as:

cosec(θ)tan(θ)+tan(θ)

Let’s factorise this expression: tan(θ)(cosec(θ)+1)

We’ll now rewrite the expression in terms of sine and cosine :

sin(θ)cos(θ)(1sin(θ)+1)

Moving the sin(θ) term inside the parentheses, we have:

1cos(θ)(1+sin(θ))

Multiplying both the denominator and numerator by (1−sin(θ)) , we have:

1−sin2(θ)cos(θ)(1−sin(θ))

Using the trigonometric identity cos2(θ)+sin2(θ)=1 *, we can rewrite this as:

cos2(θ)cos(θ)(1−sin(θ))

Dividing both the numerator and denominator by cos(θ) , we can rewrite this as:

cos(θ)1−sin(θ)

Oh dear, this is only the same as cos(theta) when 1−sin(θ)=1⇒sin(θ)=0⇒θ=180k∘ , where k is any integer

So, your equation is not true for all values of θ and thus can’t be proven true.

Instead, what we have achieved is to find the values of θ for which the equation is true; this is known as solving the equation.

However, I note that with a minor modification, I can come up with an equation that is true for all values of θ :

cosec theta/(cot theta + tan theta) =cos theta

This relies on another mathematical convention; calculations within parentheses have priority over other calculations. Example 4/(1 + 3) = 4/4 = 1.

I won’t bother to prove this, as other people has already done so.