Question

If cot θ = 7/8 , evaluate
$\frac{(1 \: + \: \sin \theta)(1 \: - \: \sin \theta) }{(1 \: + \: \cos \theta) (1 \: - \: \cos \theta) }$
$\cot^{2} \: \theta$
​

Answer 1

EXPLANATION.

cot∅ = 7/8.

As we know that,

cot∅ = 7/8 = Base/perpendicular.

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

As we know that,

Formula of :

⇒ x² - y² = (x + y)(x - y).

Using this formula in equation, we get.

$\sf \implies \dfrac{(1 - sin^{2} \theta)}{(1 - cos^{2} \theta)} = \dfrac{cos^{2} \theta}{sin^{2} \theta}$

⇒ cot²∅.

Put the value of cot∅ = 7/8 in equation, we get.

⇒ cot²∅ = (7/8)².

⇒ cot²∅ = 49/64.

                                                                                                                     

MORE INFORMATION.

Fundamental trigonometric identities.

(1) = sin²∅ + cos²∅ = 1.

(2) = 1 + tan²∅ = sec²∅.

(3) = 1 + cot²∅ = cosec²∅.

Answer 2

Given : $\bf \cot \theta = \dfrac{7}{8}$

Exigency To Evaluate : $\sf \dfrac{(1 \: + \: \sin \theta)(1 \: - \: \sin \theta) }{(1 \: + \: \cos \theta) (1 \: - \: \cos \theta) }$

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Given that ,

• $\sf \cot \theta = \dfrac{7}{8}$

❍ Basic Formulas of Trigonometry is given by :

$\begin{gathered}\boxed { \begin{array}{c c} \\ \dag \qquad \large {\underline {\bf{ Some \:Basic\:Formulas \:For\:Trigonometry \::}}}\\\\ \sf{ In \:a \:Right \:Angled \: Triangle-:} \\\\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered}$

Then,

• $\sf \cot \theta = \dfrac{7}{8} \:or\:\dfrac{Base}{Perpendicular}$

Therefore ,

• Base of Right angled triangle is 7
• Perpendicular of Right Angled Triangle is 8

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Now ,

• $\bf \dag\:\:\dfrac{(1 \: + \: \sin \theta)(1 \: - \: \sin \theta) }{(1 \: + \: \cos \theta) (1 \: - \: \cos \theta) }$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Solving \: the \: Given \: \::}}$

$:\implies \:\:\sf \dag\:\:\dfrac{(1 \: + \: \sin \theta)(1 \: - \: \sin \theta) }{(1 \: + \: \cos \theta) (1 \: - \: \cos \theta) }$

As , We know that ,

• Algebraic Identity = a² - b² = (a + b ) ( a - b )

$:\implies \:\:\sf \:\:\dfrac{(1 \: -\: \sin^2 \theta) }{(1 \: - \: \cos^2 \theta) }$

As , We know that ,

• $1 - \sin^2 \theta = \cos^2 \theta$

And ,

• $1 - \cos^2 \theta = \sin^2 \theta$

$:\implies \:\:\sf \dfrac{(\: \cos^2 \theta) }{( \: \sin^2 \theta) }$

As , We know that ,

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

$:\implies \:\:\sf \dfrac{(\: \cos^2 \theta) }{( \: \sin^2 \theta) } \:\: = \:\: \cot ^2 \theta$

As , We know that ,

• $\cot^2 \theta =\bigg( \dfrac{Base }{Perpendicular }\bigg) ^2$

Here,

• Base of Right angled triangle is 7
• Perpendicular of Right Angled Triangle is 8

Then ,

• $:\implies \:\:\sf \:\: \cot ^2 \theta = \bigg( \dfrac{7}{8}\bigg)^2$

$:\implies \:\:\bf \:\: \cot ^2 \theta = \bigg( \dfrac{49}{64}\bigg)\:\:\longrightarrow \:AnswEr$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\boxed{\begin{array}{cc} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{array}}$

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