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Question Number 215696 by BaliramKumar last updated on 15/Jan/25

Answered by MATHEMATICSAM last updated on 15/Jan/25

$${\sec }^4\mathrm{x}-{\mathrm{cosec}}^4\mathrm{x}-{2\sec }^2\mathrm{x}+{2\mathrm{cosec}}^2\mathrm{x}=\frac{63}{8}$$$$\Rightarrow {\sec }^{2}x({\sec }^{2}x-2)+{\mathrm{cosec}}^{2}x(2-{\mathrm{cosec}}^2\mathrm{x})=\frac{63}{8}$$$$\Rightarrow (1+{\tan }^{2}x)({\tan }^{2}x-1)+(1+{\cot }^{2}x)(1-{\cot }^{2}x)=\frac{63}{8}$$$$\Rightarrow ({\tan }^{4}x-1)+(1-{\cot }^{4}x)=\frac{63}{8}$$$$\Rightarrow {\tan }^{4}x-1+1-{\cot }^{4}x=\frac{63}{8}$$$$\Rightarrow {\tan }^{4}x-{\cot }^{4}x=\frac{63}{8}$$

Commented by MATHEMATICSAM last updated on 15/Jan/25

$$\mathrm{I}\mathrm{got}\mathrm{something}\mathrm{like}\mathrm{this}$$

Answered by A5T last updated on 15/Jan/25

$${\sec }^4\mathrm{x}-{\mathrm{cosec}}^4\mathrm{x}-{2\sec }^2\mathrm{x}+{2\mathrm{cosec}}^2\mathrm{x}=\frac{63}{8}$$$$\Rightarrow {({\sec }^2\mathrm{x}-1)}^2-{({\mathrm{cosec}}^2\mathrm{x}-1)}^2=\frac{63}{8}$$$${\sec }^2\mathrm{x}={\tan }^2\mathrm{x}+1;{\mathrm{cosec}}^2\mathrm{x}=1+{\cot }^2\mathrm{x}$$$$\Rightarrow {\tan }^4\mathrm{x}-\frac{1}{{\tan }^4\mathrm{x}}=\frac{63}{8}\Rightarrow {8\tan }^8\mathrm{x}-{63\tan }^4\mathrm{x}-8=0$$$${\tan }^4\mathrm{x}=\frac{63\underset{-}{+}\sqrt{{63}^2-4(-8)8}}{16}>0\Rightarrow {\tan }^4\mathrm{x}=\frac{63+65}{16}=8$$$$\Rightarrow {\tan }^2\mathrm{x}=2\sqrt{2}\Rightarrow {\cot }^2\mathrm{x}=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$$$$\Rightarrow {\tan }^2\mathrm{x}+{\cot }^2\mathrm{x}=\frac{9\sqrt{2}}{4}=\frac{9}{2\sqrt{2}}$$

Commented by BaliramKumar last updated on 15/Jan/25

$$also{tan}^{4}x-{cot}^{4}x=\frac{63}{8}$$$${tan}^{4}x-\frac{1}{{tan}^{4}x}=8-\frac{1}{8}$$$${tan}^{4}x=8$$$${tan}^{2}x=\sqrt{8}(\sqrt{8}+\frac{1}{\sqrt{8}}=\frac{9}{2\sqrt{2}})$$$$$$

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