Question

if cot tita= 7/8 then fing the value of (1+cos tita) (1-cos tita) /(sin tita +1) (sin tita -1)​

Answer 1

$\huge\sf\underline{AnsweR}:$

$\large\sf\implies-\frac{64}{105}$

$\huge\sf\underline{Solution}:$

$\large\sf{Given}:$

$\:\:\:\:\:\:\;\:\;\:\star Cos\theta =\frac{7}{8}$

$\large\sf{To\:Find}:$

$\large\red{\bigstar}\frac{(1+cos \theta) (1-cos \theta)}{(sin \theta +1) (sin \theta -1)}$

we know that,

• sinθ = p/h

• cosθ = b/h

• tanθ = p/b

• cotθ = b/p

• cscθ = h/p

• secθ = h/b

where ,

• b = base, • p = perpendicular, • h = hypotenuse

so, It is given that cosθ = 7/8

Means, base is 7 and perpendicular is 8.

• Now we have to find hypotenuse,

we can find it using Pythagoras theorem.

• h²=b²+p²

$\sf \implies h^2 = 7^2 + 8^2 \\ \implies h^2 = 49+64 \\ \implies h^2 = 113 \\ \boxed{\sf h = \sqrt{113}}$

$\bigstar \sf sin\theta = \frac{perp.}{hyp.} \\ \\ sin\theta = \frac{8}{\sqrt{113}} ……(i)$

Now,

$\large\frac{(1+cos \theta) (1-cos \theta)}{(sin \theta +1) (sin \theta -1)}$

we know,

$\\ \bf \red{(a+b)(a-b)=a^2-b^2}$

Applying this Identity here, we got

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

$\color{red}\boxed{\sf sin^2\theta+cos^2\theta= 1}$

So,

$\color{red} \sf \boxed{\sf sin^2 =1-cos^2\theta}$

$\\ \implies\sf \frac{ (sin^2 \theta)}{ (sin^2 \theta -1)}$

Put the value of sinθ that we got in (i) ...

$\\ \large \implies \sf \frac{(\frac{8}{\sqrt{113}})^2}{(\frac{8}{\sqrt{113}})^2-1}$

$\large\sf\implies \frac{\frac{64}{113}}{\frac{64}{113}-1} \\ \\ \large\sf\implies \frac{\frac{64}{\cancel{113}}}{\frac{64-113}{\cancel{113}}} \\\\\large\implies\sf \frac{64}{64-113}\\ \\ \large\implies\sf\frac{64}{-105}$

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