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Verify the following trigonometric identities: a) (1 - sin x)/cos x = 1/(sec x + tan x) b) tan x cot x = 1 c) sin z / csc z + cos z / sec z = 1

Mathematics

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5 parts
a

Verified\text{Verified}

b

Verified\text{Verified}

c

Verified\text{Verified}

d

Verified\text{Verified}

e

Verified\text{Verified}

Step-by-step solution

20 steps

Here are the verifications for each identity:

Part a· 5 steps

Verify 1−sin⁡xcos⁡x=1sec⁡x+tan⁡x\frac{1 - \sin x}{\cos x} = \frac{1}{\sec x + \tan x}

  1. Start with the left-hand side (LHS) and rewrite it using fundamental identities.

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

  2. Express in terms of sec⁡x\sec x and tan⁡x\tan x.

    LHS=sec⁡x−tan⁡x\text{LHS} = \sec x - \tan x

  3. Multiply the expression by sec⁡x+tan⁡xsec⁡x+tan⁡x\frac{\sec x + \tan x}{\sec x + \tan x}.

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

  4. Use the difference of squares formula (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2 in the numerator.

    LHS=sec⁡2x−tan⁡2xsec⁡x+tan⁡x\text{LHS} = \frac{\sec^2 x - \tan^2 x}{\sec x + \tan x}

  5. Apply the Pythagorean identity sec⁡2x−tan⁡2x=1\sec^2 x - \tan^2 x = 1.

    LHS=1sec⁡x+tan⁡x\text{LHS} = \frac{1}{\sec x + \tan x}

    This matches the right-hand side (RHS).
    Therefore, the identity is verified.
    Verified\boxed{\text{Verified}}

Part b

Verified\text{Verified}

Verify tan⁡xcot⁡x=1\tan x \cot x = 1

  1. Rewrite cot⁡x\cot x in terms of tan⁡x\tan x.

    cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x}

  2. Substitute this into the left-hand side (LHS).

    LHS=tan⁡x(1tan⁡x)\text{LHS} = \tan x \left(\frac{1}{\tan x}\right)

  3. Simplify the expression.

    LHS=1\text{LHS} = 1

    This matches the right-hand side (RHS).
    Therefore, the identity is verified.
    Verified\boxed{\text{Verified}}

Part c

Verified\text{Verified}

Verify sin⁡zcsc⁡z+cos⁡zsec⁡z=1\frac{\sin z}{\csc z} + \frac{\cos z}{\sec z} = 1

  1. Rewrite csc⁡z\csc z and sec⁡z\sec z in terms of sin⁡z\sin z and cos⁡z\cos z.

    csc⁡z=1sin⁡z\csc z = \frac{1}{\sin z} sec⁡z=1cos⁡z\sec z = \frac{1}{\cos z}

  2. Substitute these into the left-hand side (LHS).

    LHS=sin⁡z1sin⁡z+cos⁡z1cos⁡z\text{LHS} = \frac{\sin z}{\frac{1}{\sin z}} + \frac{\cos z}{\frac{1}{\cos z}}

  3. Simplify each term.

    LHS=sin⁡z⋅sin⁡z+cos⁡z⋅cos⁡z\text{LHS} = \sin z \cdot \sin z + \cos z \cdot \cos z LHS=sin⁡2z+cos⁡2z\text{LHS} = \sin^2 z + \cos^2 z

  4. Apply the Pythagorean identity sin⁡2z+cos⁡2z=1\sin^2 z + \cos^2 z = 1.

    LHS=1\text{LHS} = 1

    This matches the right-hand side (RHS).
    Therefore, the identity is verified.
    Verified\boxed{\text{Verified}}

Part d

Verified\text{Verified}

Verify cos⁡2x+2sin⁡2x=1\cos 2x + 2 \sin^2 x = 1

  1. Use the double angle identity for cos⁡2x\cos 2x that involves sin⁡2x\sin^2 x.

    cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x

  2. Substitute this into the left-hand side (LHS).

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

  3. Simplify the expression.

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

    This matches the right-hand side (RHS).
    Therefore, the identity is verified.
    Verified\boxed{\text{Verified}}

Part e

Verified\text{Verified}

Verify tan⁡12x=tan⁡x1+sec⁡x\tan \frac{1}{2}x = \frac{\tan x}{1 + \sec x}

  1. Start with the right-hand side (RHS) and express tan⁡x\tan x and sec⁡x\sec x in terms of sin⁡x\sin x and cos⁡x\cos x.

    RHS=tan⁡x1+sec⁡x=sin⁡xcos⁡x1+1cos⁡x\text{RHS} = \frac{\tan x}{1 + \sec x} = \frac{\frac{\sin x}{\cos x}}{1 + \frac{1}{\cos x}}

  2. Simplify the denominator by finding a common denominator.

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

  3. Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator.

    RHS=sin⁡xcos⁡x⋅cos⁡xcos⁡x+1\text{RHS} = \frac{\sin x}{\cos x} \cdot \frac{\cos x}{\cos x + 1}

  4. Cancel out cos⁡x\cos x.

    RHS=sin⁡x1+cos⁡x\text{RHS} = \frac{\sin x}{1 + \cos x}

  5. This expression is a known half-angle identity for tan⁡x2\tan \frac{x}{2}.

    tan⁡12x=sin⁡x1+cos⁡x\tan \frac{1}{2}x = \frac{\sin x}{1 + \cos x}

    Since the RHS simplifies to sin⁡x1+cos⁡x\frac{\sin x}{1 + \cos x}, which is equal to tan⁡12x\tan \frac{1}{2}x, the identity is verified.
    Verified\boxed{\text{Verified}}

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