TMUA Official 2018 Paper 2 · Question 13

The following is an attempted proof of the conjecture:

if $\tan\theta>0$ , then $\sin\theta+\cos\theta>1$ .

Suppose $\tan\theta>0$ , so in particular $\cos\theta\ne0$ .

(I) Since $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ , then $\sin\theta\cos\theta=\tan\theta\cos^2\theta>0$ .
(II) It follows that $1+2\sin\theta\cos\theta>1$ .
(III) Therefore $\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta>1$ ,
(IV) which factorises to give $(\sin\theta+\cos\theta)^2>1$ .
(V) Therefore $\sin\theta+\cos\theta>1$ .

Which one of the following is the case?

AThe proof is correct.

BThe proof is incorrect, and the first error occurs in line (I).

CThe proof is incorrect, and the first error occurs in line (II).

DThe proof is incorrect, and the first error occurs in line (III).

EThe proof is incorrect, and the first error occurs in line (IV).

FThe proof is incorrect, and the first error occurs in line (V).

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