Prove that (tanβ‘ A + sin β‘A)/(tanβ‘ A - sin β‘A ) = (sec A + 1)/(sec A - 1)
Last updated at Nov. 1, 2019 by Teachoo
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Question 30 (OR 2nd question) Prove that (tanβ‘γπ΄ γ+γ sinγβ‘π΄)/(tanβ‘γπ΄ γβγ π ππγβ‘π΄ ) = (π ππ π΄ + 1)/(π ππ π΄ β 1) Solving LHS (tanβ‘γπ΄ γ+γ sinγβ‘π΄)/(tanβ‘γπ΄ γβγ π ππγβ‘π΄ ) Writing everything in terms of sin A and cos A = (sinβ‘γ π΄γ/cosβ‘γ π΄γ + sinβ‘γ π΄γ)/(sinβ‘γ π΄γ/cosβ‘γ π΄γ βγ sinγβ‘γ π΄γ ) = ((sinβ‘γ π΄γ + sinβ‘π΄ cosβ‘π΄)/cosβ‘γ π΄γ )/((sinβ‘γ π΄γ β sinβ‘π΄ cosβ‘π΄)/cosβ‘γ π΄γ ) = (sinβ‘γ π΄γ + sinβ‘π΄ cosβ‘π΄)/(sinβ‘γ π΄γ β sinβ‘π΄ cosβ‘π΄ ) Taking sin A common = (sinβ‘π΄ (1 + cosβ‘π΄ ))/(sinβ‘π΄ (1 β cosβ‘π΄ ) ) = ((1 + cosβ‘π΄ ))/( (1 β cosβ‘π΄ ) ) Taking cos A common = (cosβ‘γ π΄γ (1/cosβ‘π΄ + 1))/(cosβ‘γ π΄γ (1/cosβ‘π΄ β 1) ) = ( (1/cosβ‘π΄ + 1))/( (1/cosβ‘π΄ β 1) ) = (π ππ π΄ + 1)/(π ππ π΄ β 1) = RHS Hence proved
CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic
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CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic
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