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