Prove the identity 1+cos⁡θsin⁡θ+sin⁡θ1+cos⁡θ=2csc⁡θ\frac{1+\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta} = 2\csc \theta. Hence, find all the angles from 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ which satisfy the equation
1+cos⁡2θsin⁡2θ+sin⁡2θ1+cos⁡2θ=tan⁡75∘.\frac{1+\cos 2\theta}{\sin 2\theta} + \frac{\sin 2\theta}{1+\cos 2\theta} = \tan 75^\circ.

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Worked solution

11(a)

1. Combine the fractions on the left-hand side into a single fraction

LHS=1+cos⁡θsin⁡θ+sin⁡θ1+cos⁡θ\text{LHS} = \frac{1+\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta}
=(1+cos⁡θ)2+sin⁡2θsin⁡θ(1+cos⁡θ)= \frac{(1+\cos \theta)^2 + \sin^2 \theta}{\sin \theta (1+\cos \theta)}

2. Expand the numerator

=1+2cos⁡θ+cos⁡2θ+sin⁡2θsin⁡θ(1+cos⁡θ)= \frac{1 + 2\cos \theta + \cos^2 \theta + \sin^2 \theta}{\sin \theta (1+\cos \theta)}

3. Apply the Pythagorean identity to simplify the numerator

Since cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1,
=1+2cos⁡θ+1sin⁡θ(1+cos⁡θ)= \frac{1 + 2\cos \theta + 1}{\sin \theta (1+\cos \theta)}
=2+2cos⁡θsin⁡θ(1+cos⁡θ)= \frac{2 + 2\cos \theta}{\sin \theta (1+\cos \theta)}

4. Factorise the numerator and simplify the fraction to reach the right-hand side

=2(1+cos⁡θ)sin⁡θ(1+cos⁡θ)= \frac{2(1 + \cos \theta)}{\sin \theta (1+\cos \theta)}
=2sin⁡θ= \frac{2}{\sin \theta}
=2csc⁡θ=RHS= 2\csc \theta = \text{RHS}

Answer

Proof completed.

11(b)

1. Apply the identity from part (a) to simplify the equation

From (a), 1+cos⁡θsin⁡θ+sin⁡θ1+cos⁡θ=2csc⁡θ\frac{1+\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta} = 2\csc \theta.
Substituting θ\theta with 2θ2\theta, the equation becomes:
2csc⁡2θ=tan⁡75∘2\csc 2\theta = \tan 75^\circ

2. Rearrange to find $\sin 2\theta$

2sin⁡2θ=tan⁡75∘\frac{2}{\sin 2\theta} = \tan 75^\circ
sin⁡2θ=2tan⁡75∘\sin 2\theta = \frac{2}{\tan 75^\circ}
sin⁡2θ=0.535898\sin 2\theta = 0.535898

3. Find the basic angle and solve for $2\theta$ in the required range

Since 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ, we have 0∘≤2θ≤720∘0^\circ \le 2\theta \le 720^\circ.
Basic angle α=sin⁡−1(0.535898)≈32.405∘\alpha = \sin^{-1}(0.535898) \approx 32.405^\circ
Since sin⁡2θ>0\sin 2\theta > 0, 2θ2\theta is in the 1st and 2nd quadrants.
2θ=32.405∘,180∘−32.405∘,360∘+32.405∘,540∘−32.405∘2\theta = 32.405^\circ, 180^\circ - 32.405^\circ, 360^\circ + 32.405^\circ, 540^\circ - 32.405^\circ
2θ=32.405∘,147.595∘,392.405∘,507.595∘2\theta = 32.405^\circ, 147.595^\circ, 392.405^\circ, 507.595^\circ

4. Solve for $\theta$ and round to 1 decimal place

θ=16.2025∘,73.7975∘,196.2025∘,253.7975∘\theta = 16.2025^\circ, 73.7975^\circ, 196.2025^\circ, 253.7975^\circ
θ=16.2∘,73.8∘,196.2∘,253.8∘ (to 1 d.p.)\theta = 16.2^\circ, 73.8^\circ, 196.2^\circ, 253.8^\circ \text{ (to 1 d.p.)}

Answer

θ=16.2∘,73.8∘,196.2∘,253.8∘\theta = 16.2^\circ, 73.8^\circ, 196.2^\circ, 253.8^\circ

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