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8. (a) Prove that $$ sec 2A + tan 2A = \frac{cos A + sin A}{cos A - sin A}; A \neq \frac{(2n + 1)\pi}{4}, n \in Z $$ (b) Hence solve, for $0 \leq \theta < 2\pi,$ $$ sec 2\theta + tan 2\theta = \frac{1}{2} $$ Give your answers to 3 decimal places. - Edexcel - A-Level Maths Pure - Question 9 - 2015 - Paper 3
Question 9
8. (a) Prove that $$ sec 2A + tan 2A = \frac{cos A + sin A}{cos A - sin A}; A \neq \frac{(2n + 1)\pi}{4}, n \in Z $$ (b) Hence solve, for $0 \leq \theta < 2\pi,$ $$ ... show full transcript
Worked Solution & Example Answer:8. (a) Prove that $$ sec 2A + tan 2A = \frac{cos A + sin A}{cos A - sin A}; A \neq \frac{(2n + 1)\pi}{4}, n \in Z $$ (b) Hence solve, for $0 \leq \theta < 2\pi,$ $$ sec 2\theta + tan 2\theta = \frac{1}{2} $$ Give your answers to 3 decimal places. - Edexcel - A-Level Maths Pure - Question 9 - 2015 - Paper 3
Step 1
Prove that sec 2A + tan 2A = \frac{cos A + sin A}{cos A - sin A}; A \neq \frac{(2n + 1)\pi}{4}, n \in Z
Answer
To prove the identity, we start with the left-hand side:
Next, we can use the double angle formulas:
Substituting these into our equation gives:
Now we can factor the denominator:
Thus, we can rewrite the left-hand side:
To simplify further, we recognize:
When we cancel the identical terms, we arrive at:
This completes the proof of the identity.
Step 2
Hence solve, for 0 ≤ θ < 2π, sec 2θ + tan 2θ = 1/2
Answer
Starting from the identity proven above, we have:
Now we cross-multiply to solve:
Using the double angle formula for cosine, we expand:
This leads us to:
Now, simplifying or manipulating for roots and factoring may yield solutions. For this equation, we can identify that:
- Rearranging terms might show relationships leading to specific angles.
- Use numerical methods or software to solve for valid between and .
Through numerical solving, we have:
Thus, the solutions are:
- .