Question

(1 - tan A) square + (1+tan A) square=2 sec square A



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Answer 1

Step-by-step explanation:

Take left hand side of the equation .

Answer 2

Question

$\sf{ {(1 - tanA)}^{2} + {(1 + tanA)}^{2} = 2 {sec}^{2} A}$

To prove

$\to \sf{L.H.S= R.H.S}$

Solution

L.H.S:-

${ \underline{ \boxed{\sf{identity \: used = ({a + b})^{2} }}}}$

$\sf : \implies{(1 + tanA)}^{2} + {(1 - tanA)}^{2} = 1 + {tan}^{2} A - 2 \: tanA + 1 + {tan}^{2} A + 2tanA$

$\sf{ : \implies \: {1 - tanA}^{2} + {1 + tanA}^{2} = 2(1 + {tan}^{2}A})$

$\sf : \implies {(1 - tanA)}^{2} + {(1 + tanA)}^{2} = 2 + 2 {tan}^{2} A$

$\sf{ \to \color{grey} \: we \: know \: that(1 + {tan}^{2} A) = {sec}^{2} A}$

$\sf{: \implies(1 + {tan}^{2} A) + {(1 - tan}^{2} A) = 2 {sec}^{2} A}$

$\sf{ \color{grey} \therefore \: R.H.S= L.H.S}$

Hence proved....

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know more !!

• sin² a + cos² a = 1

• 1+tan²a = sec² a

• cosec²a = 1 + cot² a

• Sin (90 – θ) = Cos θ

• Cos (90 – θ) = Sin θ

• Tan (90 – θ) = Cot θ

• Cot ( 90 – θ) = Tan θ

• Sec (90 – θ) = Cosec θ

• Cosec(90 – θ) = Sec θ

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