differentiate w.r.t. x
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i hope this will help you.
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✨HERE IS YOUR ANSWER ✨
√a²+x²/a²-x² × 3√a²-x²/a²+x²
✨now, put x² = a²cosQ
so, now: √a²-a²cos Q / a²-a²cos Q × 3√a²-a²cosQ/ a²+a²cos Q
= √1+cosQ/1-cosQ × 3√1-cosQ/1+cosQ
=. √2cos²Q2/2sin²Q/2× 3√sin²Q/2/cos²Q/2
= cotQ/2×3tanQ/2 = 3 ans.
✨HOPE IT HELPS U ✨
√a²+x²/a²-x² × 3√a²-x²/a²+x²
✨now, put x² = a²cosQ
so, now: √a²-a²cos Q / a²-a²cos Q × 3√a²-a²cosQ/ a²+a²cos Q
= √1+cosQ/1-cosQ × 3√1-cosQ/1+cosQ
=. √2cos²Q2/2sin²Q/2× 3√sin²Q/2/cos²Q/2
= cotQ/2×3tanQ/2 = 3 ans.
✨HOPE IT HELPS U ✨
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