Question Number 52609 by subhankar10 last updated on 10/Jan/19 | ||
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$$ \\ $$$$ \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \mathrm{A}\:+\:\mathrm{sin}\:^{\mathrm{4}} \mathrm{A}=\mathrm{1then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{A}−\mathrm{tan}^{\mathrm{4}} \mathrm{A} \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19 | ||
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$${sin}^{\mathrm{4}} {A}=\mathrm{1}−{sin}^{\mathrm{2}} {A} \\ $$$${sin}^{\mathrm{2}} {A}×{sin}^{\mathrm{2}} {A}={cos}^{\mathrm{2}} {A} \\ $$$${tan}^{\mathrm{2}} {A}={cosec}^{\mathrm{2}} {A} \\ $$$${tan}^{\mathrm{2}} {A}−{tan}^{\mathrm{4}} {A} \\ $$$${tan}^{\mathrm{2}} {A}\left(\mathrm{1}−{tan}^{\mathrm{2}} {A}\right) \\ $$$$={cosec}^{\mathrm{2}} {A}\left(\mathrm{1}−{cosec}^{\mathrm{2}} {A}\right) \\ $$$$=\frac{\mathrm{1}}{{sin}^{\mathrm{2}} {A}}\left(\frac{{sin}^{\mathrm{2}} {A}−\mathrm{1}}{{sin}^{\mathrm{2}} {A}}\right) \\ $$$$=\frac{{sin}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {A}−{sin}^{\mathrm{4}} {A}}{{sin}^{\mathrm{4}} {A}}\left[{since}\:{sin}^{\mathrm{2}} {A}+{sin}^{\mathrm{4}} {A}=\mathrm{1}\right] \\ $$$$=−\mathrm{1} \\ $$ | ||