Math, asked by TheMahakals, 1 year ago

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Answered by Anonymous

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Answered by BRAINLYADDICTOR

103

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$(s = z) \\ \tanθ \div 2 = \sqrt{1 - e \: \div 1 + e } \tanθ \div 2 \\ \tanθ \div 2 = \sqrt{1 - e \:\div 1 + e}tanθ \div 2 \\ s.o.b.s \\ \tan {}^{2}θ \div 2 =$

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