Question

if secA+tanA=a then prove that sinA=a^2-1/a^2-1​

Answer 1

Given ,

secA + tan =. a

Now , (secA + tan A)^2. = a^2 ➡️ ( 1/cosA + sinA/ cos A) ^2

= a^2

➡️ (1+ sinA / cosA )^2. = a^2

➡️ (1+sinA)^2 ➗ cos^2=a^2

➡️ (1+sinA) ( 1+ sinA) ➗

( 1-sin^2). = a^2

➡️ (1+sinA) (1+ sinA) ➗

(1+sinA) (1-sinA) = a^2

➡️. (1+sinA) ➗ (1- sinA) =a^2

➡️ 1+sinA = ( 1- sinA) ❌a^2

➡️ 1+sinA = a^2 ➖ sinA a^

➡️.sinA ➕ sinA a^2

=a^2 ➖ 1

➡️ sinA (1➕ a^2). = a^2 ➖1

➡️. sinA. = a^2 ➖1 / a^2 +1

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