Question

tan²o cos²o =1-cos²o​

Answer 1

To Prove :-

$\sf : \; \implies tan^{2} \theta \; cos^{2} \theta = 1 - cos^{2} \theta$

Proof :-

$\sf : \; \implies tan^{2} \theta \; cos^{2} \theta = 1 - cos^{2} \theta$

$\sf : \; \implies tan^{2} \theta \times cos^{2} \theta = 1 - cos^{2} \theta$

$\sf : \; \implies \dfrac{sin^{2} \theta}{ cos^{2} \theta} \times cos^{2} \theta = 1 - cos^{2} \theta$

$\sf : \; \implies sin^{2} \theta\times 1 = 1 - cos^{2} \theta$

$\sf : \; \implies sin^{2} \theta = 1 - cos^{2} \theta$

$\sf : \; \implies 1 - cos^{2} \theta = 1 - cos^{2} \theta$

$\boxed{\bf{ \bigstar \;\; LHS = RHS }}$

Hence Proved!

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More to know :-

$\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}$

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