1.

उचितअबुवासिकस्थान पर अस्वार यालगाकर उसे रिस्तकीप्पगड पर फिर सेस्थाबालिखे।बाधवापाडाभिखममातुरत​

Answer»

\bf{To\:Prove :}\:\sf\dfrac{cos\theta+<klux>1</klux>-sin\theta}{cos\theta-1+sin\theta}=\dfrac{1+sin\theta}{cos\theta}

\bf{Proof :}

:\implies\sf\ \dfrac{ \dfrac{(cos\theta+1-sin\theta)}{sin\theta}}{ \dfrac{(cos\theta-1+sin\theta)}{sin\theta}}\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{Dividing\ By\ sin\theta}}\\ \\ :\implies\sf \ \dfrac{cot\theta+csc\theta-1}{cot\theta-csc\theta+1}\\ \\ \\ :\implies\sf \ \dfrac{cot\theta+csc\theta-\big(csc^2\theta-cot^2\theta\big)}{cot\theta-csc\theta+1}\\ \\ {\scriptsize\qquad\bf{\dag}\:\:\tt{1= csc^2\theta-cot^2\theta}}

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:\implies\sf\ \dfrac{csc\theta+cot\theta-\big(csc\theta+cot\theta\big)\big(csc\theta-cos\theta \big)}{cot\theta-csc\theta+1}\\ \\ {\scriptsize\qquad\bf{\dag}\:\:\tt{\ a^2-b^2=(a+b)(a-b)}} \\ \\ :\implies\sf \ \dfrac{csc\theta+cos\theta [1-(csc\theta-cot\theta)]}{cot\theta-csc\theta+1}\\ \\ \\ :\implies\sf \ \dfrac{csc\theta+cot\theta \bcancel{[1-csc\theta+cot\theta]}}{\bcancel{cot\theta-csc\theta+1}}\\ \\ \\ :\implies\sf\ csc\theta+cot\theta\\ \\ \\ :\implies\sf \ \dfrac{1}{sin\theta}+\dfrac{cos\theta}{sin\theta}\\ \\ \\ :\implies\underline{\boxed{\sf \dfrac{1+cos\theta}{sin\theta}}}

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Answer:

Let the Numerator be N and DENOMINATOR be (n - 5) of the Fraction.

If 9 is subtracted from the numerator and 1 is ADDED to the denominator, the number becomes 7/2

\underline{\bigstar\:\textsf{According to the given Question :}}

:\implies\sf \dfrac{Numerator-9}{Denominator+1}=\dfrac{7}{2}\\\\\\:\implies\sf \dfrac{n-9}{(n - 5)+1}=\dfrac{7}{2}\\\\\\:\implies\sf \dfrac{n-9}{n - 4}=\dfrac{7}{2}\\\\\\:\implies\sf 2(n - 9) = 7(n - 4)\\\\\\:\implies\sf 2n - 18 = 7n - 28\\\\\\:\implies\sf 28 - 18 = 7n - 2n\\\\\\:\implies\sf 10 = 5n\\\\\\:\implies\sf \dfrac{10}{5} = n\\\\\\:\implies\sf n = 2

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\dag\:\underline{\boxed{\sf Original\:Fraction=\dfrac{n}{n-5} = \dfrac{2}{ - \:3} = \dfrac{ - \:2}{3} }}


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