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If theta is an angle, then 2theta + (1+)(1-) = ?

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Question 73: If θ\theta is an angle, then sec⁡2θ+(1+tan⁡θ)(1−tan⁡θ)=?\sec^2\theta + (1+\tan\theta)(1-\tan\theta) = ?

  1. Simplify the product (1+tan⁡θ)(1−tan⁡θ)(1+\tan\theta)(1-\tan\theta).

    This is a difference of squares, which follows the formula (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2.
    Here, a=1a=1 and b=tan⁡θb=\tan\theta.

    (1+tan⁡θ)(1−tan⁡θ)=12−(tan⁡θ)2=1−tan⁡2θ(1+\tan\theta)(1-\tan\theta) = 1^2 - (\tan\theta)^2 = 1 - \tan^2\theta

  2. Substitute this simplified term back into the original expression.

    The expression becomes:

    sec⁡2θ+(1−tan⁡2θ)\sec^2\theta + (1 - \tan^2\theta)

  3. Use the Pythagorean trigonometric identity.

    Recall the identity: sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta.

  4. Substitute sec⁡2θ\sec^2\theta with 1+tan⁡2θ1 + \tan^2\theta in the expression.

    (1+tan⁡2θ)+(1−tan⁡2θ)(1 + \tan^2\theta) + (1 - \tan^2\theta)

  5. Combine like terms.

    1+tan⁡2θ+1−tan⁡2θ1 + \tan^2\theta + 1 - \tan^2\theta (1+1)+(tan⁡2θ−tan⁡2θ)(1+1) + (\tan^2\theta - \tan^2\theta) 2+02 + 0 22

    The value of the expression is 2.

  6. Compare the result with the given options.

    A. -1
    B. 0
    C. 1
    D. 2

    The calculated value matches option D.

    The final answer is D. 2\boxed{\text{D. } 2}.

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