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Here are the proofs for both trigonometric identities.
i. Prove that cot2(23π−θ)cos2(π−θ)csc(2π−θ)sin2(π+θ)tan(23π+θ)=cosθ
Step 1: Simplify each trigonometric term using quadrant rules.
- sin(π+θ)=−sinθ⟹sin2(π+θ)=(−sinθ)2=sin2θ
- tan(23π+θ)=−cotθ
- cot(23π−θ)=tanθ⟹cot2(23π−θ)=tan2θ
- cos(π−θ)=−cosθ⟹cos2(π−θ)=(−cosθ)2=cos2θ
- csc(2π−θ)=−cscθ
Step 2: Substitute the simplified terms into the Left Hand Side (L.H.S.).
L.H.S.=tan2θ(cos2θ)(−cscθ)sin2θ(−cotθ)
Step 3: Simplify the expression.
L.H.S.=−tan2θcos2θcscθ−sin2θcotθ
The negative signs cancel out:
L.H.S.=tan2θcos2θcscθsin2θcotθ
Step 4: Express cotθ, tanθ, and cscθ in terms of sinθ and cosθ.
Recall that cotθ=sinθcosθ, tanθ=cosθsinθ, and cscθ=sinθ1.
L.H.S.=(cosθsinθ)2cos2θ(sinθ1)sin2θ(sinθcosθ)
L.H.S.=cos2θsin2θcos2θsinθ1sin2θsinθcosθ
Step 5: Cancel common terms.
In the numerator: sin2θsinθcosθ=sinθcosθ.
In the denominator: cos2θsin2θcos2θsinθ1=sin2θsinθ1=sinθ.
L.H.S.=sinθsinθcosθ
L.H.S.=cosθ
This is equal to the Right Hand Side (R.H.S.).
Therefore, cot2(23π−θ)cos2(π−θ)csc(2π−θ)sin2(π+θ)tan(23π+θ)=cosθ.
ii. Prove that sec(360∘−θ)sin(180∘+θ)cot(90∘−θ)cos(90∘+θ)sec(−θ)tan(180∘−θ)=−1
Step 1: Simplify each trigonometric term using quadrant rules.
- cos(90∘+θ)=−sinθ
- sec(−θ)=secθ (since sec is an even function)
- tan(180∘−θ)=−tanθ
- sec(360∘−θ)=secθ
- sin(180∘+θ)=−sinθ
- cot(90∘−θ)=tanθ
Step 2: Substitute the simplified terms into the Left Hand Side (L.H.S.).
L.H.S.=(secθ)(−sinθ)(tanθ)(−sinθ)(secθ)(−tanθ)
Step 3: Simplify the expression.
Multiply the terms in the numerator: (−sinθ)(secθ)(−tanθ)=sinθsecθtanθ.
Multiply the terms in the denominator: (secθ)(−sinθ)(tanθ)=−sinθsecθtanθ.
L.H.S.=−sinθsecθtanθsinθsecθtanθ
Step 4: Cancel common terms.
Assuming sinθ=0, secθ=0, and tanθ=0:
L.H.S.=−1
This is equal to the Right Hand Side (R.H.S.).
Therefore, sec(360∘−θ)sin(180∘+θ)cot(90∘−θ)cos(90∘+θ)sec(−θ)tan(180∘−θ)=−1.
Drop the next question.
