Question

8. If cos^2 a-sin^2 a = tan^2 ß, then prove that
tan^2 a=cos^2 B-sin^2 ß.​

Answer 1

Given :-

• cos²α + sin²α = sec²β – 1  

To Show :-

• cos²β – sin²β = tan²α  

Solution :-

➠ cos²α + sin²α = sec²β – 1  

➠ cos²α + sin²α + 1 = sec²β

➠ cos²β = 1/cos²α – sin²α + sin²α + cos²α  

➠ cos²β = 1/2cos²α  

➠ sin²β = 1 – cos²β  

➠ sin²β = 1 – 1/2cos²α

➠ cos²β – sin²β = 1/2cos²α – ( 1/2cos²α )  

➠ cos²β – sin²β = 1/2cos²α – 1 + 1/2cos²α

➠ cos²β – sin²β = 1/cos²α – 1  

➠ cos²β – sin²β  = sec²α – 1  

➠  cos²β – sin²β = tan²α

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• Hence Proved  !

More to know :-

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

Note : View the Table from https://brainly.in/question/41861328