Question

Find the value of p if 3cosec2A(1+COSA)(1 - COSA) =p​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$1. \: \: \: \sf{ (1 - { \cos}^{2} \theta) = { \sin}^{2} \theta \: }$

$2. \: \sf{ \cosec \theta \: \times \sin \theta \: = 1 \: }$

GIVEN

$\sf{ 3 \: { \cosec}^{2}A(1 + \cos A) (1 - \cos A) = p\: }$

TO DETERMINE

The value of p

CALCULATION

$\sf{ 3 \: { \cosec}^{2}A(1 + \cos A) (1 - \cos A) = p\: }$

$\implies \: \sf{ 3 \: { \cosec}^{2}A (1 - {\cos}^{2} A) = p\: }$

$\implies \: \sf{ 3 \: { \cosec}^{2}A \times {\sin}^{2} A = p\: }$

$\implies \: \sf{ 3 \: {({ \cosec}A \times {\sin} A)}^{2} = p\: }$

$\displaystyle \: \implies \: \sf{ 3 \: { \bigg( \frac{1}{{\sin} A} \times {\sin}A \bigg)}^{2} = p\: }$

$\displaystyle \: \implies \: \sf{3 \times {(1)}^{2} = p\: }$

$\displaystyle \: \implies \: \sf{ p\: = 3}$

RESULT

$\boxed{ \: \displaystyle \: \: \sf{ p = 3 \: \: } \: }$

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Answer 2

We have:

$3\csc^2A(1+\cos A)(1-\cos A)=p$

We have to find the value of p.

Solution:

∴ $3\csc^2A(1+\cos A)(1-\cos A)=p$

Using the algebraic identity:

(a + b)(a - b) = $a^2-b^2$

⇒ $3\csc^2A)(1-\cos^2 A)=p$

Using the trigonometric identity:

$\sin^2 A+\cos^2 A=1$

⇒ $\sin^2 A=1-\cos^2 A$

⇒ $3\csc^2A\sin^2 A=p$

⇒ $3(\csc A\sin A)=p$

⇒ $3(1)=p$

⇒ p = 3

Thus, the value of "p is equal to 3".