Question

If sinθ+ cosθ = p and tanθ + cotθ = q, then prove that q( $p^{2}$ - 1 ) = 2[ $p^{2}$ - 2/q ]
Please help (previous question had a mistake)

Answer 1

Given :

sinθ + cosθ = p and tanθ + cotθ = q

To prove :

$q(p^{2} -1)=2[q^{2} -\frac{2}{q} ]$

Step-by-step explanation:

• The following can be proved by using following these steps.

• Let LHS be $q(p^{2} -1)$ and RHS be $2[p^{2} -\frac{2}{q} ]$  

• Substitute the value in LHS.

         tanθ + cotθ[(sinθ + cosθ)$^{2}$ - 1 ]

• Expand the brackets.

        tanθ + cotθ [$sin^{2}$θ + $cos^{2}$θ+ 2sinθcosθ - 1]

• Substitute the formula in the equation

        $sin^{2}$θ+$cos^{2}$θ=1

• The expression would be

         tanθ + cotθ [ 1 - 1 + 2sinθcosθ ]

• Cancel out the terms.
• The final LHS is 2.
• Substitute the value in RHS.

        2[ (sinθ + cosθ$)^{2}$ - $\frac{2}{q}$ ]

• From the above steps, we know that

        2[ 2sinθcosθ -$\frac{2}{q}$ ]

• After substituting tanθ + cotθ in the denominator  
• The final answer is 2.
• Therefore, LHS = RHS .

Final answer :

$q(p^{2} -1)=2[q^{2} -\frac{2}{q} ]$ is proved.