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Ask QuestionPosted by Akshaya Lanka 7 years, 2 months ago
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Posted by Pawan Soni 7 years, 2 months ago
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Sia ? 6 years, 6 months ago
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Posted by Rajubb Rajubb 5 years, 8 months ago
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Sagar Jogi 7 years, 2 months ago
Posted by Aditya Tiwari 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
LHS = {tex}\sqrt { \frac { 1 + \sin \theta } { 1 - \sin \theta } } + \sqrt { \frac { 1 - \sin \theta } { 1 + \sin \theta } }{/tex}
{tex}= \sqrt { \frac { ( 1 + \sin \theta ) } { ( 1 - \sin \theta ) } \times \frac { ( 1 + \sin \theta ) } { ( 1 + \sin \theta ) } }{/tex}+ {tex}\sqrt { \frac { ( 1 - \sin \theta ) } { ( 1 + \sin \theta ) } \times \frac { ( 1 - \sin \theta ) } { ( 1 - \sin \theta ) } }{/tex}
{tex}= \sqrt { \frac { ( 1 + \sin \theta ) ^ { 2 } } { 1 - \sin ^ { 2 } \theta } } + \sqrt { \frac { ( 1 - \sin \theta ) ^ { 2 } } { 1 - \sin ^ { 2 } \theta } }{/tex}
{tex}= \sqrt { \frac { ( 1 + \sin \theta ) ^ { 2 } } { \cos ^ { 2 } \theta } } + \sqrt { \frac { ( 1 - \sin \theta ) ^ { 2 } } { \cos ^ { 2 } \theta } }{/tex} {tex}[\because sin^2\theta+cos^2\theta=1]{/tex}
{tex}= \frac { 1 + \sin \theta } { \cos \theta } + \frac { 1 - \sin \theta } { \cos \theta }{/tex}
{tex}= \frac { 1 + \sin \theta + 1 - \sin \theta } { \cos \theta }{/tex}
{tex}= \frac { 2 } { \cos \theta }{/tex}
= {tex}2sec\theta{/tex}
= RHS
Posted by Lalit Rawat 7 years, 2 months ago
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Posted by Vasu Bhardwaj 7 years, 2 months ago
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Aayushi Gupta 7 years, 2 months ago
Yogita Ingle 7 years, 2 months ago
The given A. P. is 3, 8, 13, 18, .....
Here, a = 3
d = 8 - 3 = 5
Let the nth term f=of the A. P. be 78
Then, an = = a + (n - 1)d
78 = 3 + (n - 1) 5
78 = 3 + 5n - 5
78= 5n - 2
5n = 78 + 2 = 80
n = 80/5 = 16
Hence, 16th term of the A .P. is 78.
Neetu Yadav 7 years, 2 months ago
Hello,
solution:
as we know that nth term of an AP is
Tn = a+(n-1)d
Here in given AP
a= 3
d= 5
78 = 3+(n-1)5
78-3 = (n-1)5
75/5=n-1
15 = n-1
n = 16
Sixteenth term is 78.
hHop it helps you
Posted by Animesh Pandey 7 years, 2 months ago
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Sagar Jogi 7 years, 2 months ago
Posted by Vinay Bhardwaj 7 years, 2 months ago
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Sia ? 6 years, 6 months ago
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Posted by Vinay Bhardwaj 7 years, 2 months ago
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Shalu Gupta 7 years, 2 months ago
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Piyush Parjapati 7 years, 2 months ago
Posted by Radhika :) Mishra:} 7 years, 2 months ago
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Prajjawal Chomiyal 7 years, 2 months ago
Posted by Sunidhi Chauhan 7 years, 2 months ago
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Sunidhi Chauhan 7 years, 2 months ago
Sunidhi Chauhan 7 years, 2 months ago
Posted by Akash Rauthan 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Given: In figure, {tex}\frac{{QR}}{{QS}} = \frac{{QT}}{{PR}}{/tex}and {tex}\angle{/tex} 1 = {tex}\angle{/tex} 2
To prove: {tex}\triangle PQS \sim \triangle TQR{/tex}
Proof: In {tex}\triangle {/tex} PQR {tex}\because {/tex} {tex}\angle{/tex} 1= {tex}\angle{/tex} 2
{tex}\therefore {/tex} PR = QP (1).......[ {tex}\because {/tex} sides opposite to equal angle of a triangle are equal]
Now, {tex}\frac{{QR}}{{QS}} = \frac{{QT}}{{PR}}{/tex} ......given
{tex} \Rightarrow \frac{{QR}}{{QS}} = \frac{{QT}}{{QP}}{/tex} (2).......Using(1)
Again in {tex}\triangle PQS{/tex} and {tex}\triangle TQR{/tex}
{tex}\because \frac{{QR}}{{QS}} = \frac{{QT}}{{QP}}...........From(2){/tex}
{tex}\therefore \frac{{QS}}{{QR}} = \frac{{QP}}{{QT}}{/tex} and {tex}\angle SQP = \angle RQT{/tex}
{tex}\therefore \triangle PQS \sim \triangle TQR{/tex}...........SAS similarity criterion
Posted by Ayush Burman 7 years, 2 months ago
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Posted by Susmita Sarkar 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let AP : PB = k : 1
{tex}\therefore{/tex} {tex}\frac{6 \mathrm{k}+2}{\mathrm{k}+1}{/tex} = 4
{tex}\Rightarrow{/tex} k = 1, ratio is 1 : 1
Hence m = {tex}\frac{-3+3}{2}{/tex} = 0
Posted by Vaishnavi Gajare 7 years, 2 months ago
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Kusum Panday 7 years, 2 months ago
Posted by Sanjay Bishwakarma 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
If the squared difference of the zeroes of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144, then ,we have to find the value of p.
Let {tex}\alpha{/tex} and {tex}\beta{/tex} be the zeroes of the given quadratic polynomial.
{tex}\therefore{/tex} {tex}\alpha{/tex} + {tex}\beta{/tex} = - p and {tex}\alpha\beta{/tex}= 45 ...(i)
Given, ({tex}\alpha{/tex} - {tex}\beta{/tex})2 = 144
or, ({tex}\alpha{/tex} + {tex}\beta{/tex})2- 4{tex}\alpha{/tex} {tex}\beta{/tex} = 144
or, (-p)2 - 4 {tex}\times{/tex} 45 = 144 [Using (i)]
p2 - 180 = 144
p2 = 144 + 180 = 324
{tex}\therefore{/tex} p = ± {tex}\sqrt{324}{/tex}= ± 18
Hence, the value of p is ± 18.
Posted by Abhishek Nagar 7 years, 2 months ago
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Posted by Shivam Soni 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
f(x) = x4 + 4x2 + 6
= (x2)2 + 4x2 + 6
Let x2 =n,
Then, f(x) = n2 + 4n + 6,
Here a=1,b=4,c=6
The discriminant(D) = {tex}\text{b}^2-4\mathrm{ac}=\;(4)^2-4\times1\times6=16-24=-8{/tex}
Since the discriminant is negative so this polynomial has no zeros
Hence, f(x) = x4 + 4x2 + 6 has no zero.
Posted by Arin Kumawat 7 years, 2 months ago
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Yogita Ingle 7 years, 2 months ago
A polynomial is a mathematical expression that consists of variables and constants combined using addition, subtraction and multiplication. Variables may have non-negative integer exponents. Although division by a constant is allowed, division by a variable is not allowed.
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