Question

-tanθcot(90°-θ)+secθcosec(90°-θ)+sin^2 35°+sin^2 55°/tan10°tan20°tan30°tan70°tan80°​

Answer 1

Answer:

Step-by-step explanation:

- TanθCot(90°-θ)+SecθCosec(90°-θ)+Sin²35° + Sin²55/ Tan10Tan20Tan30Tan70Tan80

We know that

//Cot(90 - θ) = Tanθ,  //Cosec(90 - θ) = Secθ, //Sin55 = Sin(90-35) = Cos35,

//Tan10 = Tan(90 -80) = Cot80, //Tan20 = Tan (90 -70) = Cot70

Substitute these values in the given problem:

=>   -Tanθ * Tanθ + Secθ*Secθ + Sin²35 + Cos²35/   Cot80Cot70Tan30Tan70Tan80

=> -Tan²θ + Sec²θ + Sin²35 + Cos²35/Cot80Cot70Tan30Tan70Tan80

We know the identities:

Sec²θ -Tan²θ = 1

Sin²θ + Cos²θ = 1

TanθCotθ = 1

=> 1 + 1 / 1 * 1 * Tan30

=> 2/(1/√3)

=> 2√3.

Answer 2

Answer:

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Step-by-step explanation:

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