Example 13 - Chapter 3 Class 11 Trigonometric Functions
Last updated at April 16, 2024 by Teachoo
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Chapter 3 Class 11 Trigonometric Functions
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Transcript
Example 13 Prove that sinβ‘γ(π₯+π¦) γ/sinβ‘γ(π₯+π¦) γ = tanβ‘γπ₯ + tanβ‘π¦ γ/tanβ‘γπ₯ βtanβ‘π¦ γ Taking L.H.S. sinβ‘γ(π₯+π¦) γ/sinβ‘γ(π₯+π¦) γ = (π¬π’π§β‘γ π γ ππ¨π¬β‘γπ + ππ¨π¬β‘γπ π¬π’π§β‘γπ γ γ γ)/π¬π’π§β‘γπ ππ¨π¬β‘γπ β ππ¨π¬β‘γπ π¬π’π§β‘π γ γ γ Dividing the numerator and denominator by πππβ‘π πππβ‘π, = ((sinβ‘π₯ cosβ‘γπ¦ + cosβ‘γπ₯ sinβ‘π¦ γ γ)/cosβ‘γπ₯ cosβ‘π¦ γ )/((sinβ‘π₯ cosβ‘γπ¦ β cosβ‘γπ₯ sinβ‘π¦ γ γ)/cosβ‘γπ₯ cosβ‘π¦ γ ) = ((sinβ‘π₯ cosβ‘π¦)/(cosβ‘π₯ cosβ‘π¦ ) + γcos π₯ sinγβ‘γπ¦ γ/(cosβ‘π₯ cosβ‘π¦ ))/(sinβ‘γπ₯ cosβ‘π¦ γ/(cosβ‘π₯ cosβ‘π¦ ) β γπππ π₯ sinγβ‘π¦/γcos π₯ cosγβ‘π¦ ) = (sinβ‘π₯/cosβ‘π₯ + sinβ‘γπ¦ γ/cosβ‘π₯ )/(sinβ‘π₯/(cosβ‘π₯ ) β sinβ‘π¦/cosβ‘π¦ ) = πππ§β‘γπ + πππ§β‘π γ/πππ§β‘γπ β πππ§β‘π γ
