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Q no.1 Prove that : (SinA + Cosec A) ² + (Cos A + Sec A) ² = 7 + Cot ²A + Tan²A

Answer :

LHS = (Sin²A + Cosec²A + 2 SinA x CosecA) + (Cos²A + Sec²A + 2 CosA x SecA)
1 1
= (Sin²A + Cosec²A + 2 SinA x SinA ) + (Cos²A + Sec²A + 2 CosA x SecA )

= (Sin²A + Cosec²A + 2) + (Cos²A + Sec²A + 2)

= (Sin²A + Cos²A) + (Cosec²A + Sec²A) + 4

= 1 + 4 + Cosec²A + Sec²A therefore { Sin²A + Cos²A = 1}

= 5 + 1+Cot²A + 1+Tan²A therefore { Cosec²A = 1+Cot²A}

and { Sec²A = 1+Tan²A }

= 7 + Cot ²A + Tan²A = RHS.

Q no.2 Find the value of x : Sin (x - 20⁰) = Cos (3x - 10⁰)

Answer :

Sin (x - 20⁰) = Sin (90⁰ - 3x + 10⁰) therefore Sin Ө⁰ = Cos (90⁰- Ө⁰)

x - 20⁰ = 90⁰ - 3x + 10⁰

3x + x = 90⁰ + 10⁰ + 20⁰

4x = 120⁰

X = 30⁰

Q no.3 If TanA = CotB then, Prove that A + B = 90⁰

Answer : TanA = CotB

TanA = Tan (90⁰ - B) therefore Cot Ө⁰ = Tan (90⁰- Ө⁰)

A = 90⁰ - B

Hence, A + B = 90⁰