Prove that cos 2x − 1 cos 2x + 1 = cosec x + cotx Show that cos 2x − 1 cos 2x + 1 = -tan x If tan = c, derive…

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Prove that cos 2x − 1 cos 2x + 1 = cosec x + cotx Show that cos 2x − 1 cos 2x + 1 = -tan x If tan = c, derive the expressions for cos A and sin A in terms of c. Hence, or otherwise, prove that if sec A - tan A = x Then, tan = x

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Step 1: Start with the left-hand side (LHS) of the equation. LHS = √((1+cos x)/(1−cos x)) Step 2: Multiply the numerator and denominator inside the square root by the conjugate of the denominator, which is (1+cos x). LHS = √((1+cos x)/(1−cos x) × (1+cos x)/(1+cos x)); LHS = √(((1+cos x)²)/(1² − cos² x)); Step 3: Apply the Pythagorean identity sin² x + cos² x = 1, which implies 1 − cos² x = sin² x. LHS = √(((1+cos x)²)/(sin² x)) Step 4: Take the square root of the numerator and the denominator. Assuming 1+cos x ≥ 0 and sin x > 0 for the principal value, we can remove the absolute values. LHS = (1+cos x)/(sin x) Step 5: Separate the fraction into two terms. LHS = (1)/(sin x) + (cos x)/(sin x); Step 6: Use the definitions of cosecant (csc x = (1)/(sin x)) and cotangent (cot x = (cos x)/(sin x)). LHS = csc x + cot x This matches the right-hand side (RHS) of the equation. Therefore, we have proven that √((1+cos x)/(1−cos x)) = csc x + cot x.

Prove that cos 2x - 1 cos 2x + 1 = cosec x + cotx Show that cos 2x - 1 cos 2x + 1 = -tan x If tan = c, derive the expressions for cos A and sin A in terms of c. Hence, or otherwise, prove that if sec A - tan A = x Then, tan = x

Answer

Step-by-step solution

Start with the left-hand side (LHS) of the equation.

Multiply the numerator and denominator inside the square root by the conjugate of the denominator, which is $(1+\cos x)$ .

Apply the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ , which implies $1 - \cos^2 x = \sin^2 x$ .

Take the square root of the numerator and the denominator.

Separate the fraction into two terms.

Use the definitions of cosecant ( $\csc x = \frac{1}{\sin x}$ ) and cotangent ( $\cot x = \frac{\cos x}{\sin x}$ ).

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