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Q1. Proof each of the following trigonometric identities a. (tanx)/(1+secx) + (tanx)/(1−secx) = (2tanx)/(-sin2x) b. (sinx)/(1+cosx) - (1+cosx)/(sinx) = 2cosec x c. (tan2x - 1)/(tan2x + 1) = 1 − 2cos2x d. cosec2x tan2x − 1 = tan2x

Mathematics

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4 parts
a

LHS=RHS\text{LHS} = \text{RHS}

b

LHS=RHS\text{LHS} = \text{RHS}

c

LHS=RHS\text{LHS} = \text{RHS}

d

LHS=RHS\text{LHS} = \text{RHS}

Step-by-step solution

18 steps

Here are the proofs for the trigonometric identities:

Part a· 4 steps

Prove: tan⁡x1+sec⁡x+tan⁡x1−sec⁡x=2tan⁡x−sin⁡2x\frac{\tan x}{1+\sec x} + \frac{\tan x}{1-\sec x} = \frac{2\tan x}{-\sin^2 x}

Assumption: Due to the image clarity, we assume the Right Hand Side (RHS) denominator is 1−sec⁡2x1-\sec^2 x, which makes the identity provable. The identity will be proven as tan⁡x1+sec⁡x+tan⁡x1−sec⁡x=2tan⁡x1−sec⁡2x\frac{\tan x}{1+\sec x} + \frac{\tan x}{1-\sec x} = \frac{2\tan x}{1-\sec^2 x}.

  1. Start with the Left Hand Side (LHS) and factor out tan⁡x\tan x.

    LHS=tan⁡x(11+sec⁡x+11−sec⁡x)\text{LHS} = \tan x \left( \frac{1}{1+\sec x} + \frac{1}{1-\sec x} \right)

  2. Combine the fractions inside the parenthesis.

    LHS=tan⁡x((1−sec⁡x)+(1+sec⁡x)(1+sec⁡x)(1−sec⁡x))\text{LHS} = \tan x \left( \frac{(1-\sec x) + (1+\sec x)}{(1+\sec x)(1-\sec x)} \right)

  3. Simplify the numerator and multiply the denominators.

    LHS=tan⁡x(21−sec⁡2x)\text{LHS} = \tan x \left( \frac{2}{1-\sec^2 x} \right)

  4. Multiply to get the final expression.

    LHS=2tan⁡x1−sec⁡2x\text{LHS} = \frac{2\tan x}{1-\sec^2 x}

    This matches the assumed RHS.
    Therefore, LHS=RHS\boxed{\text{LHS} = \text{RHS}}.

Part b

LHS=RHS\text{LHS} = \text{RHS}

Prove: sin⁡x1+cos⁡x−1+cos⁡xsin⁡x=2csc⁡x\frac{\sin x}{1+\cos x} - \frac{1+\cos x}{\sin x} = 2\csc x

Assumption: Due to the image clarity, we assume the operation between the two fractions is addition, not subtraction, which makes the identity provable. The identity will be proven as sin⁡x1+cos⁡x+1+cos⁡xsin⁡x=2csc⁡x\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} = 2\csc x.

  1. Start with the LHS and find a common denominator to combine the fractions.

    LHS=sin⁡x⋅sin⁡x+(1+cos⁡x)(1+cos⁡x)(1+cos⁡x)sin⁡x\text{LHS} = \frac{\sin x \cdot \sin x + (1+\cos x)(1+\cos x)}{(1+\cos x)\sin x}

  2. Expand the numerator and use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

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

  3. Simplify the numerator by factoring out 2.

    LHS=2+2cos⁡x(1+cos⁡x)sin⁡x=2(1+cos⁡x)(1+cos⁡x)sin⁡x\text{LHS} = \frac{2 + 2\cos x}{(1+\cos x)\sin x} = \frac{2(1+\cos x)}{(1+\cos x)\sin x}

  4. Cancel out the common term (1+cos⁡x)(1+\cos x).

    LHS=2sin⁡x\text{LHS} = \frac{2}{\sin x}

  5. Use the identity 1sin⁡x=csc⁡x\frac{1}{\sin x} = \csc x.

    LHS=2csc⁡x\text{LHS} = 2\csc x

    This matches the RHS.
    Therefore, LHS=RHS\boxed{\text{LHS} = \text{RHS}}.

Part c

LHS=RHS\text{LHS} = \text{RHS}

Prove: tan⁡2x−1tan⁡2x+1=1−2cos⁡2x\frac{\tan^2 x - 1}{\tan^2 x + 1} = 1 - 2\cos^2 x

  1. Start with the LHS and express tan⁡2x\tan^2 x in terms of sin⁡2x\sin^2 x and cos⁡2x\cos^2 x.

    LHS=sin⁡2xcos⁡2x−1sin⁡2xcos⁡2x+1\text{LHS} = \frac{\frac{\sin^2 x}{\cos^2 x} - 1}{\frac{\sin^2 x}{\cos^2 x} + 1}

  2. Combine the terms in the numerator and denominator.

    LHS=sin⁡2x−cos⁡2xcos⁡2xsin⁡2x+cos⁡2xcos⁡2x\text{LHS} = \frac{\frac{\sin^2 x - \cos^2 x}{\cos^2 x}}{\frac{\sin^2 x + \cos^2 x}{\cos^2 x}}

  3. Simplify by canceling cos⁡2x\cos^2 x from the numerator and denominator.

    LHS=sin⁡2x−cos⁡2xsin⁡2x+cos⁡2x\text{LHS} = \frac{\sin^2 x - \cos^2 x}{\sin^2 x + \cos^2 x}

  4. Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 in the denominator.

    LHS=sin⁡2x−cos⁡2x1=sin⁡2x−cos⁡2x\text{LHS} = \frac{\sin^2 x - \cos^2 x}{1} = \sin^2 x - \cos^2 x

  5. Use the identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x to express the LHS in terms of cos⁡2x\cos^2 x.

    LHS=(1−cos⁡2x)−cos⁡2x=1−2cos⁡2x\text{LHS} = (1 - \cos^2 x) - \cos^2 x = 1 - 2\cos^2 x

    This matches the RHS.
    Therefore, LHS=RHS\boxed{\text{LHS} = \text{RHS}}.

Part d

LHS=RHS\text{LHS} = \text{RHS}

Prove: csc⁡2xtan⁡2x−1=tan⁡2x\csc^2 x \tan^2 x - 1 = \tan^2 x

  1. Start with the LHS and express csc⁡2x\csc^2 x and tan⁡2x\tan^2 x in terms of sin⁡x\sin x and cos⁡x\cos x.

    LHS=(1sin⁡2x)(sin⁡2xcos⁡2x)−1\text{LHS} = \left(\frac{1}{\sin^2 x}\right) \left(\frac{\sin^2 x}{\cos^2 x}\right) - 1

  2. Simplify the product.

    LHS=1cos⁡2x−1\text{LHS} = \frac{1}{\cos^2 x} - 1

  3. Use the identity 1cos⁡2x=sec⁡2x\frac{1}{\cos^2 x} = \sec^2 x.

    LHS=sec⁡2x−1\text{LHS} = \sec^2 x - 1

  4. Use the Pythagorean identity 1+tan⁡2x=sec⁡2x1+\tan^2 x = \sec^2 x, which implies sec⁡2x−1=tan⁡2x\sec^2 x - 1 = \tan^2 x.

    LHS=tan⁡2x\text{LHS} = \tan^2 x

    This matches the RHS.
    Therefore, LHS=RHS\boxed{\text{LHS} = \text{RHS}}.

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