(B) If $\cot \theta = \frac{7}{8}$ , then find the value of $\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$ .

Q: Why does the answer come out to $\cot^2\theta$ and not $\cot \theta$?

Because both numerator ($\cos^2\theta$) and denominator ($\sin^2\theta$) are squared quantities. Their ratio is $(\cos \theta/\sin \theta)^2 = \cot^2\theta$, so you must square the given value: $(7/8)^2 = \frac{49}{64}$.

Q: Which identity is essential here?

The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$, rearranged as $1 - \sin^2\theta = \cos^2\theta$ and $1 - \cos^2\theta = \sin^2\theta$.

Introduction to Trigonometry

Trigonometric identities

Official Answer

The value of the expression equals $\cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}$ . It is a clean rational number because the messy $\sqrt{113}$ hypotenuse from a triangle method would have cancelled out anyway.

cot θ7/8difference of squares1 − sin²θ = cos²θ1 − cos²θ = sin²θcot²θ49/64Pythagorean identity

Marking Scheme

11 mark: simplifying numerator and denominator to $(1 - \sin^2\theta)/(1 - \cos^2\theta)$ and applying $\sin^2\theta + \cos^2\theta = 1$ to get $\cos^2\theta/\sin^2\theta$ .
21 mark: identifying the expression as $\cot^2\theta$ and substituting $\cot \theta = \frac{7}{8}$ to obtain $\frac{49}{64}$ .
3Full marks for the equivalent triangle method if it correctly reaches $\frac{49}{64}$ ; deduct for leaving the answer as $\cot^2\theta$ without substituting the value.

Hint

Recognise $(1+x)(1-x) = 1 - x^2$ ; then use $1 - \sin^2\theta = \cos^2\theta$ and $1 - \cos^2\theta = \sin^2\theta$ to reduce the whole thing to $\cot^2\theta$ .

Quick Oral Answer

I use $(1+a)(1-a) = 1 - a^2$ , so the fraction becomes $(1 - \sin^2\theta)/(1 - \cos^2\theta) = \cos^2\theta/\sin^2\theta = \cot^2\theta$ ; with $\cot \theta = \frac{7}{8}$ , that is $\left(\frac{7}{8}\right)^2 = \frac{49}{64}$ .

Analysis & Explanation

Recognise the algebraic identity $(1 + a)(1 - a) = 1 - a^2$ hidden in the expression before touching any numbers.

Concept

Numerator $(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2\theta$ ; denominator $(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2\theta$ .
The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ converts these to cos²θ and sin²θ respectively.
The ratio $\cos^2\theta/\sin^2\theta$ is exactly $\cot^2\theta$ , so the whole expression equals $\cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}$ .

Common mistakes

Trying to find θ or building a triangle (base $7$ , perpendicular $8$ , hypotenuse $\sqrt{113}$ ) — much longer and invites errors with an irrational hypotenuse.
Stopping at $\cot^2\theta$ without substituting the given value $\frac{7}{8}$ .

Real-world

Reducing a messy expression to a single known quantity before plugging in numbers mirrors how engineers simplify formulas before substituting measured values.

Common Mistakes

1Getting the ratio upside down as $\tan^2\theta = \frac{64}{49}$ by mixing up which of $1 - \sin^2\theta$ and $1 - \cos^2\theta$ equals $\cos^2\theta$ versus $\sin^2\theta$ .
2Forgetting to square the given ratio and writing the answer as $\frac{7}{8}$ instead of $\left(\frac{7}{8}\right)^2 = \frac{49}{64}$ .
3Wasting time building a triangle with irrational hypotenuse $\sqrt{113}$ and making arithmetic slips, instead of simplifying symbolically to $\cot^2\theta$ first.

Interesting Facts

The identity $\sin^2\theta + \cos^2\theta = 1$ is a direct restatement of the Pythagoras theorem on a right triangle whose hypotenuse is 1, which is why it works for every angle.

The expression looks complicated but collapses to a single ratio — a reminder that in trigonometry, spotting the difference-of-squares pattern $(1+a)(1-a)$ can turn a four-factor expression into a one-line answer.

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Frequently Asked Questions

Why does the answer come out to $\cot^2\theta$ and not $\cot \theta$ ?

Because both numerator ( $\cos^2\theta$ ) and denominator ( $\sin^2\theta$ ) are squared quantities. Their ratio is $(\cos \theta/\sin \theta)^2 = \cot^2\theta$ , so you must square the given value: $(7/8)^2 = \frac{49}{64}$ .

Can I solve it by making a triangle?

Yes. With $\cot \theta = \frac{7}{8}$ take base 7, perpendicular 8, hypotenuse $\sqrt{49+64} = \sqrt{113}$ . Then $\cos^2\theta/\sin^2\theta = \frac{49}{113} \div \frac{64}{113} = \frac{49}{64}$ . It works but is longer and error-prone, so the identity method is preferred.

Which identity is essential here?

The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ , rearranged as $1 - \sin^2\theta = \cos^2\theta$ and $1 - \cos^2\theta = \sin^2\theta$ .

(B) If cot⁡θ=78\cot \theta = \frac{7}{8}cotθ=87​, then find the value of (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}(1+cosθ)(1−cosθ)(1+sinθ)(1−sinθ)​.