Math, asked by sm779528, 9 months ago

Solve the following problems on paper/ notebook.
Q.1 Prove that cosh^7x= [cosh 7x + 7 cosh 5x + 21 cosh 3x + 35 coshx] ​

Answers

Answered by sourasghotekar123
0

Answer:

cosh^7x= [cosh 7x + 7 cosh 5x + 21 cosh 3x + 35 coshx] is proved

Step-by-step explanation:

From the above question:

cosh^7x= [cosh 7x + 7 cosh 5x + 21 cosh 3x + 35 coshx]

sine of ∠A = sin A = Side opposite to ∠A/ Hypotenuse

cosine of ∠A = cos A = Side adjacent to ∠A/ Hypotenuse

tangent of ∠A = tan A = (Side opposite to ∠A)/ (Side adjacent to ∠A)

cosecant of ∠A = cosec A = 1/sin A = Hypotenuse/ Side opposite to ∠A

secant of ∠A = sec A = 1/cos A = Hypotenuse/ Side adjacent to ∠A

cotangent of ∠A = cot A = 1/tan A = (Side adjacent to ∠A)/ (Side opposite to ∠A)

Also, tan A = sin A/cos A

         cot A = cos A/sin A

         sec2θ – tan2θ = 1

         (sec θ – tan θ)(sec θ + tan θ) = 1

         (sec θ – tan θ) l = 1 {from (i)}

         sec θ – tan θ = 1/l….(ii)

Adding (i) and (ii),

         tan θ + sec θ + sec θ – tan θ = l + (1/l)

         2 sec θ = (l2 + 1)l

         sec θ = (l2 + 1)/2l

LHS = RHS

Hence, cosh^7x= [cosh 7x + 7 cosh 5x + 21 cosh 3x + 35 coshx] ​is proved.

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