Prove that $$sec^2 x - cosec^2 x = tan^2 x - cot^2 x.$$ (ii) Given that $$y = arccos x, \, -1 ≤ x ≤ 1 \, and \, 0 ≤ y ≤ \, π,$$ (a) express arcsin x in terms of y. (b) Hence evaluate arccos x + arcsin x. Give your answer in terms of π.

A-Level Edexcel Maths Pure Differentiation: Prove that $$sec^2 x

Prove that $$sec^2 x - cosec^2 x = tan^2 x - cot^2 x.$$ (ii) Given that $$y = arccos x, \, -1 ≤ x ≤ 1 \, and \, 0 ≤ y ≤ \, π,$$ (a) express arcsin x in terms of y - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 4

Question 2

$Prove-that--$$sec^2-x---cosec^2-x-=-tan^2-x---cot^2-x.$$----(ii)-Given-that----$$y-=-arccos-x,-\,--1-≤-x-≤-1-\,-and-\,-0-≤-y-≤-\,-π,$$----(a)-express-arcsin-x-in-terms-of-y-Edexcel-A-Level Maths Pure-Question 2-2006-Paper 4.png$

Prove that $$sec^2 x - cosec^2 x = tan^2 x - cot^2 x.$$ (ii) Given that $$y = arccos x, \, -1 ≤ x ≤ 1 \, and \, 0 ≤ y ≤ \, π,$$ (a) express arcsin x in ter... show full transcript

Worked Solution & Example Answer:Prove that $$sec^2 x - cosec^2 x = tan^2 x - cot^2 x.$$ (ii) Given that $$y = arccos x, \, -1 ≤ x ≤ 1 \, and \, 0 ≤ y ≤ \, π,$$ (a) express arcsin x in terms of y - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 4

Step 1

Prove that $sec^2 x - cosec^2 x = tan^2 x - cot^2 x$

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Answer

To prove this identity, we start with the left-hand side:
$LHS = sec^2 x - cosec^2 x.$
Using the definitions of secant and cosecant:
$sec^2 x = \frac{1}{cos^2 x}, \, cosec^2 x = \frac{1}{sin^2 x},$
we rewrite it as:
$LHS = \frac{1}{cos^2 x} - \frac{1}{sin^2 x} = \frac{sin^2 x - cos^2 x}{sin^2 x cos^2 x}.$
Now we apply the identity for tangent and cotangent:
$tan^2 x = \frac{sin^2 x}{cos^2 x}, \, cot^2 x = \frac{cos^2 x}{sin^2 x}.$
Rearranging gives us:
$RHS = tan^2 x - cot^2 x = \frac{sin^2 x}{cos^2 x} - \frac{cos^2 x}{sin^2 x} = \frac{sin^4 x - cos^4 x}{sin^2 x cos^2 x}.$
Thus, both sides are equal, confirming the identity.

Step 2

Given that $y = arccos x$, $-1 ≤ x ≤ 1$, and $0 ≤ y ≤ π$, express arcsin x in terms of y.

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Answer

Since we know $y = arccos x$ , we can express $x$ in terms of $y$ as follows:
$x = cos y.$
Next, using the Pythagorean identity, we can find arcsin x:
$arcsin x = arcsin(cos y) = \frac{π}{2} - y.$
This gives us the expression for arcsin x in terms of y.

Step 3

Hence evaluate arccos x + arcsin x. Give your answer in terms of π.

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101 rated

Answer

Now we'll evaluate $arccos x + arcsin x$ :
$arccos x = y, \, arcsin x = \frac{π}{2} - y.$
Combining these two gives:
$arccos x + arcsin x = y + \left(\frac{π}{2} - y\right) = \frac{π}{2}.$
Thus, the final answer is:
$arccos x + arcsin x = \frac{π}{2}.$

Prove that $$sec^2 x - cosec^2 x = tan^2 x - cot^2 x.$$ (ii) Given that $$y = arccos x, \, -1 ≤ x ≤ 1 \, and \, 0 ≤ y ≤ \, π,$$ (a) express arcsin x in terms of y - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 4

Worked Solution & Example Answer:Prove that $$sec^2 x - cosec^2 x = tan^2 x - cot^2 x.$$ (ii) Given that $$y = arccos x, \, -1 ≤ x ≤ 1 \, and \, 0 ≤ y ≤ \, π,$$ (a) express arcsin x in terms of y - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 4

Prove that $sec^2 x - cosec^2 x = tan^2 x - cot^2 x$

Given that $y = arccos x$, $-1 ≤ x ≤ 1$, and $0 ≤ y ≤ π$, express arcsin x in terms of y.

Hence evaluate arccos x + arcsin x. Give your answer in terms of π.

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