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Gandhan Pavar Gandhan Pavar 1 year, 10 months ago
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Preeti Dabral 1 year, 10 months ago
{tex}\begin{aligned} & \text { To prove : } \frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sec ^{12} \mathrm{~A}}{1-\cos \mathrm{A}} \\ & \text { LHS } \frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{1+\frac{1}{\cos \mathrm{A}}}{\frac{1}{\cos \mathrm{A}}} \\ & =\frac{\frac{\cos \mathrm{A}+1}{\cos \mathrm{A}}}{\frac{1}{\cos \mathrm{A}}} \\ & =\frac{\cos \mathrm{A}+1}{\cos \mathrm{A}} \times \frac{\cos \mathrm{A}}{1} \\ & =1+\cos \mathrm{A} \\ & =1+\cos \mathrm{A} \times \frac{1-\cos \mathrm{A}}{1-\cos \mathrm{A}} \\ & =\frac{(1)^2-(\cos \mathrm{A})^2}{1-\cos \mathrm{A}} \\ & =\frac{1-\cos 2 \mathrm{~A}}{1-\cos \mathrm{A}} \end{aligned}{/tex}
{tex}\begin{aligned} & =\frac{\sin ^2 \mathrm{~A}}{1-\cos \mathrm{A}} \\ & =\mathrm{RHS} \end{aligned}{/tex}
Posted by Ashutosh Singh 1 year, 10 months ago
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Preeti Dabral 1 year, 10 months ago
Given quadratic equation is p(x-4)(x-2)+(x-1)² = 0
=> p(x²-4x-2x+8)+(x²-2x+1) = 0
=> p(x²-6x+8)+(x²-2x+1) = 0
=> px²-6px+8p+x²-2x+1 = 0
=> (px²+x²)+(-6px-2x)+8p+1 = 0
=> (p+1)x²-(6p+2)x+(8p+1) = 0
On comparing with the standard quadratic equation is ax²+bx+c = 0 then
a = p+1
b = -(6p+2)
c = 8p+1
Given that
Given equation has real and equal roots
=> The discriminant = 0
We know that
The discriminant of ax²+bx+c = 0 is b²-4ac
Therefore, b²-4ac = 0
=> [-(6p+2)]²-4(p+1)(8p+1) = 0
=> (6p+2)²-4(p+1)(8p+1) = 0
=> (36p²+24p+4)-4(8p²+p+8p+1) = 0
=> (36p²+24p+4)-4(8p²+9p+1) = 0
=> 36p²+24p+4-32p²-36p-4 = 0
=> (36p²-32p²)+(24p-36p)+(4-4) = 0
=> 4p² -12p+0 = 0
=> 4p² - 12 p = 0
=> 4p(p-3) = 0
=> 4p = 0 or p-3 = 0
=> p = 0/4 or p = 3
=> p = 0 or p = 3
If p = 0 the term containing p doesn't exist .
Therefore, p = 3
Posted by Anuradha Kumari 1 year, 10 months ago
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𝕷𝖊𝖌𝖊𝖓𝖉 🏆 🍑 🎀 𝒜𝓁☯𝓃𝑒 𝑅𝒶𝓈𝒽𝒾 🎀 🍑 1 year, 10 months ago
Posted by Deepika Chowdary 1 year, 10 months ago
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Preeti Dabral 1 year, 10 months ago
To find the value of k
sol: A quadratic equation has equal real roots if the discriminant b2−4ac=0
x2−kx+4=0 is of form ax2+bx+c=0
∴a=1,b=−k,c=4
Hence the discriminant is
(−k)2−4(1)(4)=0⟹k2−16=0⟹k2=16
Therefore, k=±4
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Pratiksha Yadav 1 year, 10 months ago
1Thank You