\sqrt{1-sin thita /1+sin thita = sec thita -tan thita
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LHS = √ [ 1 - sin∅ / 1 + sin∅ ]
=> √ [ (1 - sin∅)(1 - sin∅) / (1 + sin∅) (1 - sin∅) ]
=> √ ( 1 - sin∅ )² / √ (1²-sin²∅)
=> 1 - sin ∅ / √ ( 1 - sin²∅)
=> 1 - sin ∅ / √ cos²∅
[ sin²∅ + cos²∅ = 1
=> cos²∅ = 1 - sin²∅ ]
=> 1 - sin∅ / cos∅
=> 1/ cos∅ - sin∅/cos∅
=> sec ∅ - tan∅
since , LHS = RHS
hence , PROVED
hope this helps
=> √ [ (1 - sin∅)(1 - sin∅) / (1 + sin∅) (1 - sin∅) ]
=> √ ( 1 - sin∅ )² / √ (1²-sin²∅)
=> 1 - sin ∅ / √ ( 1 - sin²∅)
=> 1 - sin ∅ / √ cos²∅
[ sin²∅ + cos²∅ = 1
=> cos²∅ = 1 - sin²∅ ]
=> 1 - sin∅ / cos∅
=> 1/ cos∅ - sin∅/cos∅
=> sec ∅ - tan∅
since , LHS = RHS
hence , PROVED
hope this helps
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