Question

★ Trigonometry ★

1: If $\sf{cot\:\theta \:=\: \dfrac{7}{8}}$ , evaluate :

• $\boxed{\sf{\red{\dfrac{( 1 \:+\: sin \theta ) ( 1 \:-\: sin \theta )}{( 1 \:+\: cos \theta ) ( 1 \:-\: cos \theta )} }}}$

• $\boxed{\sf{\red{cot² \theta }}}$

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Answer 1

Question:

$\cos \alpha = \frac{7}{8} then \: \frac{(1 + \sin\alpha)(1 - \sin\alpha ) }{(1 + \cos \alpha )( 1 - \cos\alpha ) } =$

Solution:

$here \: tan \alpha = \frac{1}{cot \alpha } \\ tan \alpha = \frac{1}{ \frac{7}{8 } } \\ tan \alpha = \frac{8}{7}$

$then \: by \: using \: phythagoraen \: \: triplet \: find \: sin \alpha \: and \: cos \alpha$

$we \: get \: the \: third \: side \: of \: the \: triangle(hypotenuse) \: is \: \sqrt{113}$

$\sin( \alpha ) = \frac{8}{ \sqrt{113} }$

$\cos( \alpha ) = \frac{7}{ \sqrt{113} }$

$\frac{(1 + sin \alpha )(1 - \sin \alpha) }{ (1 + cos \alpha)(1 - \cos \alpha ) } \\ it \: is \: in \: the \: form \: of \: (a + b)(a - b)$

it is a²-b²

$\frac{ {1}^{2} - { \sin }^{2} \alpha }{ {1}^{2} - \ { \cos }^{2} \alpha }$

$\frac{ {1}- \frac{ {8}^{2} }{ { \sqrt{(113}) }^{2} } }{1 - \frac{ {7}^{2} }{ \sqrt{ {(113)}^{2} } } }$

$\frac{1 - \frac{64}{113} }{1 - \frac{49}{113} }$

$\frac{49}{64}$

Answer:

$\frac{49}{64}$

More information:

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

Answer 2

Before solving:-

Here, we can see trigonometric identities being used. Usually, identities are used for fast calculations. First, let's simplify the fraction and let's apply it.

What's used:-

Trigonometric Identity $\sin^2\theta+\cos^2\theta=1$

Solution:-

Evaluating the first value

In the first view of trigonometric calculations, we expect them to be boring and messy. But it is actually not when we use a trick. We can calculate in a very simple way. The trick is seeing that,

• $1-\sin^2\theta=\cos^2\theta$
• $1-\cos^2\theta=\sin^2\theta$

Hence, given fraction

$=\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$

$=\dfrac{1-\sin^2\theta}{1-\cos^2\theta}$

$=\dfrac{\cos^2\theta}{\sin^2\theta}$

$=\cot^2\theta=\boxed{\dfrac{49}{64} }$.

Evaluating the second value

When we write the power of a term we write as $(\cot\theta)^2$. However, in trigonometric ratios, it is possible to write it as $\cot^2\theta$, but it is only possible for natural power.

So, $\cot^2\theta=(\cot\theta)^2$.

We can substitute the given value.

Evaluated value

$=\cot^2\theta$

$=(\cot\theta)^2$

$=(\dfrac{7}{8} )^2=\boxed{\dfrac{49}{64} }$

More information

Kinds of reciprocal trigonometric functions.

• $\sin\theta\rightarrow \csc\theta$
• $\cos\theta\rightarrow \sec\theta$
• $\tan\theta\rightarrow\cot\theta$

Inverse trigonometric functions.

→ These are used in finding the angles when you know the trigonometric values. Substituting trigonometric values in these functions gives the angles.

• $\sin\theta\rightarrow \arcsin$
• $\cos\theta\rightarrow \arccos$
• $\tan\theta\rightarrow\arctan$