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Question: If \(\tan\theta = t,\) then \(\tan 2\theta + \sec 2\theta =\)...

If tan⁡θ=t,\tan\theta = t, then tan⁡2θ+sec⁡2θ=\tan 2\theta + \sec 2\theta =

A

1+t1−t\frac{1 + t}{1 - t}

B

1−t1+t\frac{1 - t}{1 + t}

C

2t1−t\frac{2t}{1 - t}

D

2t1+t\frac{2t}{1 + t}

Answer

1+t1−t\frac{1 + t}{1 - t}

Explanation

Solution

tan⁡2θ=2tan⁡θ1−tan⁡2θ,cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\tan 2\theta = \frac{2\tan\theta}{1 - \tan^{2}\theta},\cos 2\theta = \frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta}

tan⁡2θ+sec⁡2θ=2t1−t2+1+t21−t2=(1+t)2(1−t)(1+t)=1+t1−t\tan 2\theta + \sec 2\theta = \frac{2t}{1 - t^{2}} + \frac{1 + t^{2}}{1 - t^{2}} = \frac{(1 + t)^{2}}{(1 - t)(1 + t)} = \frac{1 + t}{1 - t}.