Question

If x=a secA cosB,y=b secA sinB and z=c tanA;show that:

$\frac{x^2}{a^2} +\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$

Class 10 trigonometry no spams any brainly star answer it!

Answer 1

Answer:

Hey mate here is your answer:

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Solution:

→ $L.H.S=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2} =1,$,

→$\frac{a\:secA\:cosB}{a^2}+\frac{b\:secA\:sinB}{b^2}-\frac{c\:tanA^2}{c^2}$,

→$\frac{a^2\:sec^2A\:cos^2B}{a^2} +\frac{b^2\:sec^2A\:sin^2B}{b^2}-\frac{c^2\:tan^2A}{c^2}$,

→$sec^2A\:cos^2B+sec^2A\:sin^2B-tan^2A$,

→$sec^2A(cos^2B+sin^2B)-tan^2A$,

→$sec^2A-tan^2A$                                  [∵cos²B+sin²B=1],

→ 1.                                                  [∵sec²A-tan²A=1].

Hence L.H.S=R.H.S.

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$\boxed{\text{Hope it helps you}}$

Answer 2

Answer: Do u need thx mate ??