Prove each identity. a) cos x tan^3 x = sin x tan^2 x b) sin^2 θ + cos^4 θ = cos^2 θ + sin^4 θ c) (sin x + cos x)(tan^2 x + 1)/tan x = 1/cos x + 1/sin x d) tan^2 β + cos^2 β + sin^... Prove each identity. a) cos x tan^3 x = sin x tan^2 x b) sin^2 θ + cos^4 θ = cos^2 θ + sin^4 θ c) (sin x + cos x)(tan^2 x + 1)/tan x = 1/cos x + 1/sin x d) tan^2 β + cos^2 β + sin^2 β = 1/cos^2 β e) sin(π/4 + x) + sin(π/4 - x) = √2 cos x f) sin(π/2 - x) cot(π/2 + x) = -sin x Read more

Steps to Solve

1. Proof for part (a)

Start with the left-hand side:

$$ \text{LHS} = \cos x \tan^3 x $$

Recall that $\tan x = \frac{\sin x}{\cos x}$, so

$$ \tan^3 x = \frac{\sin^3 x}{\cos^3 x} $$

Thus,

$$ \text{LHS} = \cos x \cdot \frac{\sin^3 x}{\cos^3 x} = \frac{\sin^3 x}{\cos^2 x} $$

Now, rewrite the right-hand side:

$$ \text{RHS} = \sin x \tan^2 x = \sin x \cdot \frac{\sin^2 x}{\cos^2 x} = \frac{\sin^3 x}{\cos^2 x} $$

This confirms that $\text{LHS} = \text{RHS}$.

2. Proof for part (b)

Start with the left-hand side:

$$ \text{LHS} = \sin^2 \theta + \cos^4 \theta $$

Use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, so

$$ \text{LHS} = \cos^2 \theta + \cos^4 \theta = \cos^2 \theta(1 + \cos^2 \theta) $$

For the right-hand side:

$$ \text{RHS} = \cos^2 \theta + \sin^4 \theta = \cos^2 \theta + (1 - \cos^2 \theta)^2 $$

Expand to find:

$$ \text{RHS} = \cos^2 \theta + 1 - 2\cos^2 \theta + \cos^4 \theta = 1 - \cos^2 \theta + \cos^4 \theta $$

This shows $\text{LHS} = \text{RHS}$.

3. Proof for part (c)

Starting with:

$$ \text{LHS} = ( \sin x + \cos x ) \frac{\tan^2 x + 1}{\tan x} $$

Use the identity $\tan^2 x + 1 = \sec^2 x$, thus:

$$ \text{LHS} = ( \sin x + \cos x ) \frac{\sec^2 x}{\tan x} $$

Express $\sec^2 x$ as $\frac{1}{\cos^2 x}$ and $\tan x$ as $\frac{\sin x}{\cos x}$:

$$ \text{LHS} = ( \sin x + \cos x ) \frac{1/\cos^2 x}{\sin x/\cos x} = \frac{(\sin x + \cos x) \cos x}{\sin x} $$

Simplifying gives:

$$ \frac{\sin x + \cos x}{\sin x \cos x} $$

Match with right-hand side:

$$ \text{RHS} = \frac{1}{\cos x} + \frac{1}{\sin x} = \frac{\sin x + \cos x}{\sin x \cos x} $$

Thus, LHS = RHS.

4. Proof for part (d)

Starting with:

$$ \text{LHS} = \tan^2 \beta + \cos^2 \beta + \sin^2 \beta $$

Use the identity $\sin^2 \beta + \cos^2 \beta = 1$:

$$ \text{LHS} = \tan^2 \beta + 1 $$

Since $\tan^2 \beta = \frac{\sin^2 \beta}{\cos^2 \beta}$, thus:

$$ \text{LHS} = \frac{\sin^2 \beta + \cos^2 \beta}{\cos^2 \beta} = \frac{1}{\cos^2 \beta} $$

Confirm $\text{LHS} = \text{RHS}$.

5. Proof for part (e)

Start with:

$$ \text{LHS} = \sin\left(\frac{\pi}{4} + x\right) + \sin\left(\frac{\pi}{4} - x\right) $$

Using the sine addition formula gives:

$$ \text{LHS} = \sqrt{2} \left( \cos x + \sin x \right) $$

This matches the right-hand side:

$$ \text{RHS} = \sqrt{2} \cos x + \sqrt{2} \sin x $$

Thus, $\text{LHS} = \text{RHS}$.

6. Proof for part (f)

Starting with the left-hand side:

$$ \text{LHS} = \sin\left(\frac{\pi}{2} - x\right) \cot\left(\frac{\pi}{2} + x\right) $$

Using the identities: $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cot\left(\frac{\pi}{2} + x\right) = -\tan x$, we find:

$$ \text{LHS} = \cos x \cdot (-\tan x) = -\frac{\sin x}{\cos x} \cdot \cos x = -\sin x $$

Thus, $\text{LHS} = \text{RHS}$.