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Ask QuestionPosted by Shubham Lohar 7 years, 4 months ago
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Posted by Rupal ???? 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
LHS ={tex}\sqrt { \frac { \sec \theta - 1 } { \sec \theta + 1 } } + \sqrt { \frac { \sec \theta + 1 } { \sec \theta - 1 } }{/tex}
Rationalise the denominator and we get,
{tex}= \sqrt{\frac{(sec\theta-1)^2}{sec^2\theta-1}}+ \sqrt{\frac{(sec\theta+1)^2}{sec^2\theta-1}}{/tex}
= {tex}\frac { ( \sec \theta - 1 ) + ( \sec \theta + 1 ) } { \sqrt { ( \sec \theta + 1 ) ( \sec \theta - 1 ) } }{/tex}
= {tex}\frac { 2 \sec \theta } { \sqrt { \sec ^ { 2 } \theta -1 } } = \frac { 2 \sec \theta } { \sqrt { \tan ^ { 2 } \theta } } = \frac { 2 \sec \theta } { \tan \theta }{/tex}
= {tex}2 \times \frac { 1 } { \cos \theta } \times \frac { \cos \theta } { \sin \theta }{/tex}
= {tex}2 \times \frac { 1 } { \sin \theta }{/tex}
= 2 cosec{tex}\theta{/tex}
= RHS
Hence Proved
Posted by Hariom Thakur 7 years, 4 months ago
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Posted by Riddhi Siddhi 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Given: cosec A + cot A = m
{tex}\Rightarrow{/tex} (cosec A + cot A)2 = (m)2 [squaring both sides ]
{tex}\Rightarrow{/tex} cosec2A + cot2A + 2 cosec A cot A = m2 .......(1)
Now, LHS
{tex}=\frac{m^{2}-1}{m^{2}+1}{/tex}
{tex}=\frac{\ cosec ^{2} A+\cot ^{2} A+2 \ cosec A \cot A-1}{\ cosec ^{2} A+\cot ^{2} A+2 \ cosce A \cdot \cot A+1}{/tex}. [ From (1) ]
{tex}=\frac{\cot ^{2} A+\cot ^{2} A+2 \ cosec A \cdot \cot A}{\ cosec ^{2} A+\ cosec ^{2} A+2 \ cosec A \cdot \cot A}{/tex} [Since, Cosec2A - Cot2A = 1]
{tex}=\frac{2 \cot ^{2} A+2 \ cosec A \cot A}{2 \ cosec ^{2} A+2 \ cosec A \cot A}{/tex}
{tex}=\frac{2 \cot A(\cot A+\ cosec A)}{2 \ cosec A(\ cosec A+\cot A)}{/tex}
{tex}=\frac{\cot A}{cosec A}{/tex}
{tex}=\frac{\frac{\cos A}{\sin A}}{\frac{1}{\sin A}}{/tex}
{tex}=\frac{\cos A}{\sin A} \times \frac{\sin A}{1}{/tex}
= cos A = RHS
Hence, Proved.
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Sia ? 6 years, 5 months ago
{tex}2x^2 - 8x - m = 0{/tex}
{tex}\because{/tex}{tex}\frac{5}{2}{/tex} is one root
2({tex}\frac{5}{2}{/tex})2 - 8{tex}\times{/tex} {tex}\frac{5}{2}{/tex} {tex}- m = 0{/tex} {tex}\Rightarrow{/tex} {tex}\frac{2 \times 25}{4}{/tex} {tex}- 20 - m = 0{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{25 - 40}{2}{/tex} {tex}- m = 0{/tex} {tex}\Rightarrow{/tex} {tex}\frac{-15}{2}{/tex} = m {tex}\Rightarrow{/tex}m = {tex}\frac{-15}{2}{/tex}.
Also, sum of roots = -{tex}\frac{b}{a}{/tex} = {tex}\frac{-(-8)}{2}{/tex} = 4
One root = {tex}\frac{5}{2}{/tex} [given]
{tex}\therefore{/tex} Other root = 4 - {tex}\frac{5}{2}{/tex} = {tex}\frac{3}{2}{/tex}
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Sia ? 6 years, 5 months ago
Let the speed of the boat in still water be 'x' km/hr and speed of the stream be 'y' km/
Speed = Distance / Time
{tex}\therefore{/tex} {tex}\frac { 30 } { x - y } + \frac { 28 } { x + y } = 7{/tex}
and {tex}\frac { 21 } { x - y } + \frac { 21 } { x + y } = 5{/tex}
Let {tex}\frac { 1 } { x - y } \text { be } a \text { and } \frac { 1 } { x + y } \text { be } b{/tex}
30a + 28b = 7 ......(i)
21a + 21b = 5 ......(ii)
Multiplying (i) by 3 and (ii) by 4 and then subtracting.
{tex}90a+84b=21{/tex} ..............(iii)
{tex}84a+84b=20 {/tex} ..............(iv)
By solving (iii) and (iv)
{tex}90a-21=84a-20{/tex}
{tex}\Rightarrow{/tex}6a= 1
{tex}\Rightarrow{/tex} {tex}a = \frac { 1 } { 6 }{/tex}
Putting this value of ,a in eqn., (i),
{tex}30 \times \frac { 1 } { 6 } + 28 b = 7{/tex}
{tex}28 b = 7 - 30 \times \frac { 1 } { 6 } = 2{/tex}
{tex}\therefore{/tex}{tex}b = \frac { 1 } { 14 }{/tex}
x + y = 14 ...(iv)
Now, {tex}a = \frac { 1 } { x - y } = \frac { 1 } { 6 }{/tex}
{tex}\Rightarrow{/tex} x - y = 6
{tex}\Rightarrow{/tex}x = y + 6 .....(v)
Putting (iv) in (v)
y + 6 + y = 14
{tex}\Rightarrow{/tex} y = 4
Hence, speed of the boat in still water = 10 km/hr and speed of the stream = 4 km/hr.
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