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If a +b =1 and a^{2} ^{2} -b^{2}^{2} =5, than a^{2}

Question

If $a sec θ + b tan θ = 1$ and $a^{2} sec^{2} θ - b^{2} tan^{2} θ$ = $5$ , than $a^{2} b^{2} + 4 a^{2}$

9 b^{2}

\frac{9}{a ^{2}}

\frac{- 2}{b}

9

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Daniel Discussed

a sec θ + b tan θ = 1

and

a^{2} sec^{2} θ - b^{2} tan^{2} θ

5

, than

a^{2} b^{2} + 4 a^{2}

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

a sec θ = 1 - b tan θ a^{2} sec^{2} θ = 1 + b^{2} tan^{2} θ - 2 b tan θ .......... (i)

Also,

a^{2} sec^{2} θ = 5 + b^{2} tan^{2} θ ............. (ii)

∴

(i) and (ii)

\Rightarrow 1 - 2 b tan θ = 5 \Rightarrow b tan θ = - 2 ∴ a sec θ = 3 \Rightarrow a = 3 cos θ \Rightarrow a^{2} = 9 cos^{2} θ .......... (iii) b tan θ = - 2 \Rightarrow b^{2} \frac{sin ^{2} θ}{cos ^{2} θ} = 4 \Rightarrow cos^{2} θ = \frac{b ^{2}}{b ^{2} + 4} ........ (i v)

From (iii) and (iv), we get

a^{2} = \frac{9 b ^{2}}{b ^{2} + 4} \Rightarrow (b^{2} + 4) a^{2} = b^{2} a^{2} + 4 a^{2} = 9 b^{2}

Therefore, Answer is

9 b^{2}

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Question Text	If $a sec θ + b tan θ = 1$ and $a^{2} sec^{2} θ - b^{2} tan^{2} θ$ = $5$ , than $a^{2} b^{2} + 4 a^{2}$
Answer Type	Text solution:1
Upvotes	150