Solving and proving identities

In this section we shall explore proving and solving identities. The identities explored here will include; secant θ, cosecant θ, cotangent θ and tangent θ.
For whatever purpose or applications you should be able to simplify, prove identities and solve equations involving secant θ, cosecant θ and cotangent θ. In the following examples we shall explore how to work with these identities.

Simplifying identities

Below we shall explore some examples where simplifying identities has been applied.

Example; sin θ cot θ sec θ

Suppose we had to simplify sinθcotθsecθ.
First we would need to rewrite the expression in terms of sin and cos, we know that;

…so we get;

…this simplifies to;

Example: sinθcosθ(secθ + cosecθ)

Here is another expression that we’re going to simplify; Suppose we had to simplify;

First let us look at the bracketed expression. Again we will have to rewrite the expression in terms of sin and cos, we know that;

…this gives us;

…using common denominators we get;

We have the expression as;

Now we can multiply both sides by sinθ cosθ to get;

We have managed to reduce the expression to;

Proving identities

Example: Proving cos^3>3θ

In this example we’re going to prove the following expression;
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First let us consider the left hand side (LHS) and we shall look at the denominator and numerator separately. First we shall look at the numerator, the numerator is;
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Again we shall rewrite the expression in terms of sin and cos as we saw above;
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So for the numerator we get;
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Note we shall need the numerator we have just simplified above. Now we shall simplify the denominator; The denominator is;
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Again we need to rewrite the expression in terms of sin and cos, we know that;
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…and…
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…so we get;
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…this simplifies to;
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Note that here;
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therefore we get;
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So now we can bring the numerator and denominator simplified above here to form an expression;
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Solving sec and cot

In the following examples we shall look at two examples solving sec θ and cot θ and we’re looking in the interval 0° ≤ θ ≤ 360°.
Suppose we had to solve sec θ = 2.5, we know that;
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…therefore we can rewrite the expression as;
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…simplifying it we get;
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Here we can see that cosθ is negative, that must mean that θ is in 2nd and 3rd quadrants. You can see this on the quadrant diagram below;
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The acute angle to the horizontal is cos^-1 (+0.4). So the angles that we need are;
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Reading off from the diagram we get;
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Next we shall try cot 2θ = 0.6. We know that;
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…therefore;
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So now we can work with cot 2θ. To make the process simpler we let;
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So now we’re solving for;
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Since we’re working with 2θ let us also double the interval to get an interval of 0 ≤ x ≤ 720°. We shall draw the diagram with the acute angle x which is tan^-1 5/3. Tan is positive in the 1st and 3rd quadrant;
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So we have x angles as;
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Remember above we have been solving for x but we need to find θ. Remember we let 2θ = x. That is;
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…so now;
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Using 1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ

In this section we shall explore using the identities; 1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ.

Example

First we shall prove that;
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This isn’t so obvious because it evolves from another identity, we know that;
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Let us divide both sides in the equation above with cos²θ that gives;
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You can already recognise some identities, this simplifies to;
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This gives us;
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This proves the expression.

Example

Now we shall prove the following expression;
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Again we’re going to use the identity;
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Let us divide both sides by sin²θ. This gives;
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We know that;
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So the expression becomes;
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…therefore…
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Here is another very simple workout using tan before continuing to prove more identities; The question goes;

Given that tan A = -5/12, and that angle A is obtuse, find the exact value of sec A and sin A

There are two ways we can solve this. The first method involves using the quadrant diagram; We shall use the identity;
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…we can simply replace the known value of tan in the expression above …so we get;
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…therefore sec A is;
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But remember we need to take into account that angle A is obtuse; The answer above does not take into account that; angle A is obtuse. We can use the quadrant diagram below;
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Since angle A is obtuse. In the 2nd quadrant sec A is negative, therefore;
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The second method involves using a right-angled triangle with;
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…so the triangle we want to draw is;
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Remember that;
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…we shall use this together with Pythagoras’s theorem to find sec θ; In the right-angled triangle above we can see that;
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…that must mean that;
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Since A is in the 2nd quadrant, cos A is negative and therefore sec A is negative; so the answer is;
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How about finding sin A when tan A = -5/12. This is simple as well. We know that;
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…make sin A the subject to get;
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…because cos A = 1/sec A that must mean that; cos A = -12/12, we just substitute this in the equation above together with tan A; the expression becomes;
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Now let us go back to proving identities.

Example: Prove cosec⁴ – cot⁴

First we shall prove that;
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We shall start with the left hand side (LHS)
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We can factorise here since it is a difference of two squares, so we get;
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…we know that; 1 + cot² θ ≡ cosec ² θ that must mean that cosec ² θ – cot ² θ ≡ 1, so we get;
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…we know that;
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…so this simplifies to;
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…using sin² θ + cos² θ ≡ 1 we get;
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Proving sec ² θ – cos² θ

In this example we shall show that;
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We shall start from the right hand side and write each term in terms of sin θ and cos θ. So we get;
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We know that;
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…therefore we get;
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Here we shall use cos² θ ≡ 1 – sin² θ and sec ² θ ≡ 1 + tan² θ. The hint is in the LHS since it is in terms of cos² θ and sec² θ, so we get the expression as;
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…then…
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This is the same as the left hand side expression.

Solving 4cosec² θ – 9 = cot θ

Now we shall solve the equation 4cosec² θ – 9 = cot θ in the interval 0° ≤ θ ≤ 360°. The equation 4cosec² θ – 9 = cot θ is a quadratic equation, therefore we need to rewrite it in terms of one trigonometrical function only here we shall use;
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…to get;
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Note here you can replace cot θ with x if it is too confusing this may make it simple for you for instance you could say;
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…that should work fine as well. I will leave the cot θ in the express;
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…we can factorise to get;
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…so now;
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That must mean that;
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So for tan 4/5 we get the following, you may draw the quadrant diagram if you want to as shown below;
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Note for this situation tan θ is positive in the 1st and 3rd quadrants. So the angles are;