证明以下身份(1 – 13)
问题1秒4θ -秒2θ=黄褐色4θ+黄褐色2θ
解决方案:
We have
sec4θ – sec2θ = tan4θ + tan2θ
Taking LHS
= sec4θ – sec2θ
= sec2θ(sec2θ – 1)
Using sec2 θ = tan2θ + 1, we get
= (1 + tan2θ)tan2θ
= tan2θ + tan4θ
Hence, LHS = RHS (Proved)
问题2.罪6θ+ COSθ6 = 1 – 2 3sin 2θcosθ
解决方案:
We have
sin6θ + cos6θ = 1 – 3sin2θcos2θ
Taking LHS
= sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
Using a3 + b3 = (a + b)(a2 + b2 – ab), we get
= (sin2θ + cos2θ)(sin4θ + cos4θ – sin2θcos2θ)
Using a2 + b2 = (a + b)2 – 2ab and sin2θ + cos2θ = 1, we get
= (1)[(sin2θ + cos2θ)2 – 2sin2θcos2θ – sin2θcos2θ]
= (1)[(1)2 – 3sin2θcos2θ]
= 1 – 3sin2θcos2θ
Hence, LHS = RHS (Proved)
问题3.(cosecθ–sinθ)(secθ–cosθ)(tanθ+cotθ)= 1
解决方案:
We have
(cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1
Taking LHS
= (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ)
Using cosecθ = 1/sinθ and secθ = 1/cosθ
=
=
=
= 1
Hence, LHS = RHS (Proved)
问题4cosecθ(secθ – 1) – cotθ(1 – COSθ)=tanθ – SINθ
解决方案:
We have
cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ
Taking LHS
=
=
=
=
=
=
=
Hence, LHS = RHS(Proved)
问题5
解决方案:
We have
Taking LHS
=
Using a2 – b2 = (a + b)(a – b) and a3 + b3 = (a + b)(a2 + b2ab), we get
=
=
=
=
= sinA
Hence, LHS = RHS(Proved)
问题6。
解决方案:
We have
Taking LHS
=
Using tanA = sinA/cosA and cotA = cosA/sinA, we get
=
=
=
=
Using a3 – b3 = (a – b)(a2 + b2 + ab), we get
=
=
=
Using cosecA = 1/sinA and secA = 1/cosA, we get
= secAcosecA + 1
Hence, LHS = RHS(Proved)
问题7。
解决方案:
We have
Taking LHS
=
Using a3 ± b3 = (a ± b)(a2 + b2 ± ab), we get
=
Using sin2θ + cos2θ = 1, we get
= 1 – sinAcosA + 1 + sinAcosA
= 2
Hence, LHS = RHS(Proved)
问题8.(secAsecB + tanAtanB) 2 –(secAtanB + tanAsecB) 2 = 1
解决方案:
We have
(secAsecB + tanAtanB)2 – (secAtanB + tanAsecB)2 = 1
Taking LHS
= (secAsecB + tanAtanB)2 – (secAtanB + tanAsecB)2
Expanding the above equation using the formula
(a + b)2 = a2 + b2 + 2ab
= (secAsecB)2 + (tanAtanB)2 + 2(secAsecB)(tanAtanB) –
(secAtanB)2 – (tanAsecB)2 – 2(secAtanB)(tanAsecB)
= sec2Asec2B + tan2Atan2B – sec2Atan2B – tan2Asec2B
= sec2A(sec2B – tan2B) – tan2A(sec2B – tan2B)
= sec2A – tan2A -(Using sec2θ – tan2θ = 1)
= 1
Hence, LHS = RHS(Proved)
问题9。
解决方案:
We have
Taking RHS
=
=
= ×
=
=
=
=
=
=
=
=
=
= ×
=
=
=
=
Hence, RHS = LHS(Proved)
问题10。
解决方案:
We have
Taking LHS
=
Using 1 + tan2x = sec2x and 1 + cot2x = cosec2x, we get
=
=
=
=
=
Using a2 + b2 = (a + b)2 – 2ab, we get
=
=
=
Hence, LHS = RHS (Proved)
问题11
解决方案:
We have
Taking LHS
=
By using the formulas cotθ = cosθ/sinθ and tanθ = sinθ/cosθ, we get
=
=
=
Using a3+b3 = (a + b)(a2 + b2 – ab), we get
=
=
= 1 – (sin2θ + cos2θ) + sinθcosθ
= 1 – 1 + sinθcosθ
= sinθcosθ
Hence, LHS = RHS (Proved)
问题12。
解决方案:
We have
=
Taking LHS
=
=
=
=
=
=
=
=
=
Hence, LHS = RHS(Proved)
问题13(1 +tanαtanβ)2 +(tanαtanβ)2 =秒2αsec2β
解决方案:
We have
(1 + tanαtanβ)2 + (tanα – tanβ)2 = sec^2αsec2β
Taking LHS
= (1 + tanαtanβ)2 + (tanα – tanβ)2
= (1 + tan2αtan2β + 2tanαtanβ) + (tan2α + tan2β – 2tanαtanβ)
= 1 + tan2αtan2β + tan2α + tan2β
= (1 + tan2β) + tan2α(1 + tan2β)
= (1 + tan2β)(1 + tan2α)
= sec2αsec2β
Hence, LHS = RHS (Proved)