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Posted by Aaliya Yduvanshi.... 6 years, 2 months ago
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Posted by Shivam Yadav 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
LHS: (2x–1)(x–3) =2x2–6x–x+3 = 2x2–7x+3
RHS: (x+5)(x–1) = x2–x+5x–5 = x2 + 4x – 5
Now; 2x2–7x+3 = x2+4x–5
Or, 2x2–7x+3–x2−4x+5 = 0
Or, x2–11x+8=0
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Posted by Naved Ahmad 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
From right triangle BOD
BD = √(OB²-OD²) = √(13²- 8²) = √105
from right triangle POD
PD = √(OP²-OD²) = √(13²-8²) = √105
PB = BD + PD = 2√105 PB²= 420
from right triangle APB
AP = √(AB²- PB²) =√26²- 420 = √676-420 = √256 = 16

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Suhani Hotwani 6 years, 2 months ago
Aadya Singh ? 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
{tex}L H S=\frac{\cos A-\sin A+1}{\cos A+\sin A-1}{/tex}
{tex}=\frac{\sin A(\cos A-\sin A+1)}{\sin A(\cos A+\sin A-1)}{/tex}
{tex}=\frac{\sin A \cos A-\sin ^{2} A+\sin A}{\sin A(\cos A+\sin A-1)}{/tex}
{tex}=\frac{\sin A \cos A+\sin A-\left(1-\cos ^{2} A\right)}{\sin A(\cos A+\sin A-1)}{/tex}
{tex}=\frac{\sin A(\cos A+1)-(1-\cos A)(1+\cos A)}{\sin A(\cos A+\sin A-1)}{/tex}
{tex}=\frac{(1+\cos A)(\sin A+\cos A-1)}{\sin A(\cos A+\sin A-1)}{/tex}
{tex}=\frac{(1+\cos A)(\sin A+\cos A-1)}{\sin A(\cos A+\sin A-1)}{/tex}
{tex}=\frac{1}{\sin A}+\frac{\cos A}{\sin A}{/tex}
= cos A + cot A = RHS
Proved
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Shreya... #Be Positve Be Happy 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
We know that the tangent segments from an external point to a circle are equal
{tex}\therefore{/tex} AP = AS ........(1)
BP = BQ .......(2)
CR = CQ .......(3)
DR = DS .......(4)
Adding (1), (2), (3) and (4), we get
(AP + BP) + (CR + DR) = (AS + BQ + CQ + DS)
{tex}\Rightarrow{/tex} AB + CD = (AS + DS) + (BQ + CQ)
{tex}\Rightarrow{/tex} AB + CD = AD + BC
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Yogita Ingle 6 years, 2 months ago
Nth term of the sequence is given by t_n = (n - 1)(n -2 )(n -3 ), here n is integers e.g., n = 1, 2, 3, 4, .........
First term , put n = 1
Then, t₁ = (1 - 1)(1 - 2)(1 - 3) = 0 × (-1) × (-2) = 0
2nd term, put n = 2
Then, t₂ = (2 - 1)(2 - 2)(2 - 3) = 1 × 0 × (-1) = 0
3rd term , put n = 3
Then, t₃ = (3 - 1)( 3 - 2)(3 - 3) = 2 × 1 × 0 = 0
Here it is clear that first three terms of the sequence are zero.
Now, put n = 4
Then, t₄ = (4 - 1)(4 - 2)(4 - 3) = 3 × 2 × 1 = 6 > 0 e.g., positive
Put n = 5
Then, t₅ = (5 - 1)(5 - 2)(5 - 3) = 4 × 3 × 2 = 24 > 0 e.g., positive
Hence, after 3rd term all the terms of the sequence are positive.
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