Question

Cosec theeta+cottheeta=p then find costheeta in terms of p
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Answer 1

Solution

$cosec \theta + cot \theta = p$

$\implies \dfrac{1}{sin \theta} + \dfrac{cos \theta}{sin \theta} = p$

[ Because cosecθ = 1/sinθ and cotθ = cosθ/sinθ ]

$\implies \dfrac{1 + cos \theta}{sin \theta} = p$

Squaring on both sides

$\implies \bigg( \dfrac{1 + cos \theta}{sin \theta} \bigg)^{2} = (p)^{2}$

$\implies \dfrac{(1 + cos \theta)^{2} }{sin^{2} \theta} = p^{2}$

$\implies \dfrac{ {1}^{2} + cos^{2} \theta + 2(1)(cos \theta)}{sin^{2} \theta} = p^{2}$

[ Because (x + y)² = x² + y² + 2xy ]

$\implies \dfrac{ 1 + cos^{2} \theta + 2cos \theta}{1 - cos^{2} \theta} = p^{2}$

[ Becaus sin²θ = 1 - cos²θ ]

$\implies 1 + cos^{2} \theta + 2cos \theta = {p}^{2} (1 - cos^{2} \theta )$

$\implies 1 + cos^{2} \theta + 2cos \theta = {p}^{2} - {p}^{2} cos^{2} \theta$

$\implies {p}^{2}cos^{2} \theta + cos^{2} \theta + 2cos \theta + 1 - {p}^{2} = 0$

$\implies ({p}^{2} +1) cos^{2} \theta+ 2cos \theta + 1 - {p}^{2} = 0$

Now it is in the form of a quadratic equation ax² + bx + c = 0

Comparing with ax² + bx + c = 0 we get,

• a = p² + 1
• b = 2
• c = 1 - p²

Discriminant = b² - 4ac

= 2² - 4(p² + 1)(1 - p²)

= 4 - 4[ (p² + 1){ - ( p² - 1) } ]

= 4 + 4(p² + 1)(p² - 1)

= 4 + 4(p^4 - 1)

= 4(1 + p^4 - 1)

= 4p^4

By using Quadratic formula

$\boxed{ x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

$\implies cos \theta= \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

$\implies cos \theta= \dfrac{ - 2 \pm \sqrt{4 {p}^{4} } }{2( {p}^{2} + 1) }$

$\implies cos \theta= \dfrac{ - 2 \pm \sqrt{(2 {p}^{2} )^{2} } }{2({p}^{2} + 1)}$

$\implies cos \theta= \dfrac{ - 2 \pm 2 {p}^{2} }{2({p}^{2} + 1)}$

$\implies cos \theta= \dfrac{ - 2 + 2 {p}^{2} }{2({p}^{2} + 1)} \quad or \quad cos \theta= \dfrac{ - 2 - 2 {p}^{2} }{2({p}^{2} + 1)}$

$\implies cos \theta= \dfrac{ 2(- 1 + {p}^{2}) }{2({p}^{2} + 1)} \quad or \quad cos \theta= \dfrac{2( - 1 - {p}^{2} )}{2({p}^{2} + 1)}$

$\implies cos \theta= \dfrac{{p}^{2} - 1}{{p}^{2} + 1} \quad or \quad cos \theta= \dfrac{ - ( {p}^{2} + 1) }{{p}^{2} + 1}$

$\implies cos \theta= \dfrac{{p}^{2} - 1}{{p}^{2} + 1} \quad or \quad cos \theta = - 1$

[ Ignoring cosθ = - 1 as cotθ = undefined and cosecθ = undefined ]

$\implies \boxed{ cos \theta=\dfrac{{p}^{2} - 1}{{p}^{2} + 1}}$

Hence, value of cosθ is (p² - 1)/(p² + 1).

Answer 2

Question :-

$if \: \csc \: \theta + \cot \: \theta = p \: \\ \\ \sf{ find \: \cos \: \theta \: in \: terms \: of \: p}$

Answer :-

$\boxed{ \cos \: \theta = \frac{p^{2}-1}{p^{2}+1} }$

Step - by - step explanation :-

Given that ,

$\rightarrow \: \csc \: \theta + \cot \: \theta = p \\ \\ \rightarrow \: \csc \: \theta = p - \cot \: \theta \\ \\ \sf{squaring \: on \:both \: sides} \\ \\ \rightarrow \: { \csc}^{2} \theta = {p}^{2} + { \cot}^{2} \theta - 2p \cot \: \theta \\ \\ \rightarrow \: { \csc}^{2} \theta - { \cot}^{2} \theta = {p}^{2} - 2p \cot \: \theta \\ \\ \rightarrow \: {p}^{2} - 2p \cot \: \theta = 1 \: \: \: \: \\ \\ \rightarrow \: {p}^{2} - 2p \: \bigg(\frac{ \cos \: \theta}{sin \: \theta} \bigg) = 1 \: \: ...(1) \\ \\ \sf{now \: we \: take \: } \\ \\ \rightarrow \: \csc \: \theta + \cot \: \theta = p \\ \\ \rightarrow \: \frac{ 1 + \cos \: \theta}{ \sin \: \theta} = p \\ \\ \rightarrow \: \sin \: \theta = \frac{1+ \cos \:\theta}{p} \: \: \: .....(2)$

Substitute the value from.eq.(2) in (1)

After this step ,refer to the attachment.