# Proving Trigonometric identities Worksheet[With Solution]

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Problem: 01 Prove that: √sec²θ + cosec²θ=tanθ+cotθ

Solution :

LHS= √sec²θ + cosec²θ

=√(1+tan²θ)(1+cot²θ)                  [Using identity : sec²θ=1+tan²θ and cosec²θ=1+cot²θ]

= √(tan²θ + cot²θ + 2)

= √(tan²θ + cot²θ +2tanθ.cotθ)  [tanθ.cotθ=1]

=√(tanθ+cotθ)²

=tanθ+cotθ = RHS

Hence proved

Problem: 02 Prove that : \frac{sin³θ+cos³θ}{sinθ+cosθ} +sinθcosθ=1

Solution:

LHS= \frac{sin³θ+cos³θ}{sinθ+cosθ} +sinθcosθ

=\frac{(sinθ+cosθ)(sin²θ+cos²θ-sinθcosθ)}{sinθ+cosθ} +sinθcosθ

[Using identity : sin³θ + cos³θ=(sinθ+cosθ)(sin²θ+cos²θ-sinθcosθ)

=\frac{(sinθ+cosθ)(sin²θ+cos²θ-sinθcosθ)}{sinθ+cosθ} +sinθcosθ

=(sin²θ + cos²θ-sinθcosθ ) + sinθcosθ

=(1-sinθcosθ)+( sinθcosθ)            [Using : sin²θ + cos²θ=1 ]

=1   RHS

Hence proved

Problem: 03 Prove that : \frac{tanA}{(1-cotA)}+\frac{cotA}{1-tanA}=(1+tanA+cotA)

Solution :

LHS=\frac{tanA}{(1-cotA)}+\frac{cotA}{1-tanA}

Convert tan , cot ratios in sin and cos ratios

=\frac{\frac{SinA}{CosA}}{1-\frac{CosA}{SinA}} +\frac{\frac{CosA}{SinA}}{1-\frac{SinA}{CosA}}

=\frac{sin²A}{cosA(sinA-cosA)} + \frac{cos²A}{sinA(cosA-sinA)}

= \frac{sin²A}{cosA(sinA-cosA)}\frac{cos²A}{sinA(sinA-cosA)}

=\frac{sin³A-cos³A}{sinAcosA(sinA-cosA)}

=\frac{(sinA-cosA)(sin²A+cos²A+sinAcosA)}{sinAcosA(sinA-cosA)}

[Using identity : sin³A-cos³A=(sinA-cosA)(sin²A+cos²A+sinAcosA)]

=\frac{1+sinAcosA}{sinAcosA}      [Using sin²A+cos²A=1]

Now simplify RHS

(1+tanA+cotA)=(1+\frac{sinA}{cosA}+\frac{cosA}{sinA}

(1+tanA+cotA)=((\frac{sinAcosA+sin²A+cos²A}{sinAcosA}

(1+tanA+cotA)=\frac{sinAcosA + 1 }{sinAcosA}

Hence LHS=RHS    proved

Problem: 04 Prove that : (sinθ+cosecθ)²+(cosθ+secθ)²=(7+tan²θ+cot²θ)

Solution:

LHS= (sinθ+cosecθ)²+(cosθ+secθ)²

Expand using (a+b)²=a²+b²+2ab

= (sin²θ+cosec²θ+2sinθcosecθ) + (cos²θ + sec²θ+2cosθsecθ)

= (sin²θ+cosec²θ+2) + (cos²θ + sec²θ+2)  [Since : sinθcosecθ =1 and cosθsecθ=1 ]

= (sin²θ+cos²θ)+(cosec²θ+sec²θ)+4

= (1) +(1+cot²θ+1+tan²θ)+4

=1+1 +1+4+cot²θ+tan²θ

= 7+ cot²θ+tan²θ

Problem: 05 Prove that : (cosecθ-secθ)²=\frac{1-cosθ}{1+cosθ}

Solution:

LHS = (cosecθ-secθ)²

Convert cosec and sec ration into sin and cos

=(\frac{1}{sinθ}\frac{1}{cosθ}

= (\frac{1-cosθ}{sinθ})²

= \frac{(1-cosθ)²}{sin²θ}

= \frac{(1-cosθ)(1-cosθ)}{1-cos²θ}         [Using identity : sin²θ=1-cos²θ]

= \frac{(1-cosθ)(1-cosθ)}{(1-cosθ)(1+cosθ)}

= \frac{(1-cosθ}{1+cosθ}=RHS

Hence proved

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Problem: 06 Prove that :

\frac{1}{(cosecθ-secθ)}\frac{1}{sinθ}=\frac{1}{sinθ}\frac{1}{(cosecθ+cotθ)}

Solution:

LHS = \frac{1}{(cosecθ-secθ)}\frac{1}{sinθ}

= \frac{1}{(cosecθ-secθ)}×\frac{(cosecθ+secθ)}{(cosecθ+secθ)}\frac{1}{sinθ}

=\frac{(cosecθ+cotθ)}{(cosec²θ-cot²θ}-cosecθ  [Since; \frac{1}{sinθ}=cosecθ]

= \frac{(cosecθ+cotθ)}{(cosec²θ-cot²θ)}-cosecθ

= \frac{(cosecθ+cotθ)}{(1)}-cosecθ  [Using identity : cosec²θ-cot²θ =1 ]

= cosecθ+cotθ-cosecθ

=cotθ

Similarly simplify RHS

RHS = \frac{1}{sinθ}\frac{1}{(cosecθ+cotθ)}

= \frac{1}{sinθ}\frac{1}{(cosecθ+cotθ)}×\frac{cosecθ-cotθ}{cosecθ-cotθ}

= cosecθ-\frac{(cosecθ-cotθ)}{(cosec²θ-cot²θ)}      [Using identity : cosec²θ-cot²θ =1 ]

=cosecθ – (cosecθ -cotθ)

=cosecθ-cosecθ+cotθ

=cotθ

Hence ,LHS=RHS

Proved

Problem: 07 Prove that : \frac{1-sinθ}{1+sinθ}=(secθ-tanθ)²

Solution:

LHS= \frac{1-sinθ}{1+sinθ}

= \frac{1-sinθ}{1+sinθ} ×\frac{1-sinθ}{1-sinθ}

= \frac{(1-sinθ)²}{(1+sinθ)(1-sinθ)}

= \frac{(1-sinθ)²}{(1-sin²θ)}

= \frac{(1-sinθ)²}{(cos²θ)}       [Using : 1-sin²θ=cos²θ]

=(\frac{1-sinθ}{cosθ}

= (\frac{1}{cosθ}\frac{sinθ}{cosθ}

= (secθ – tanθ)² =RHS

Hence proved

Problem:08 If 1+sin²θ=3sinθcosθ then prove that tanθ=1 or \frac{1}{2}

Solution:

Dividing LHS and RHS by cos²θ

\frac{1+sin^{2} \theta }{cos^{2} \theta } =\frac{3sin\theta cos\theta }{cos^{2} \theta }

⇒ sec²θ+tan²θ=3tanθ

⇒ (1+tan²θ) + tan²θ = 3tanθ

⇒ 2tan²θ-3tanθ+1=0

⇒ 2tan²θ -2tanθ-tanθ+1=0

⇒ 2tanθ(tanθ-1) – (tanθ-1)=0

⇒ (tanθ-1)(2tanθ-1)=0

(tanθ-1)= 0 or (2tanθ-1)=0

tanθ=1 or tanθ=\frac{1}{2}

Problem:09 \frac{√1+sinθ}{√1-sinθ} + \frac{√1-sinθ}{√1+sinθ}=2secθ

Solution

LHS = \frac{√1+sinθ}{√1-sinθ} + \frac{√1-sinθ}{√1+sinθ}

LCM of (√1-sinθ) and (√1+sinθ) is (√1-sinθ)(√1+sinθ)

= \frac{1+sinθ + 1-sinθ}{(√1-sinθ)(√1+sinθ)}

= \frac{2}{√1-sin²θ}

= \frac{2}{√cos²θ}

= \frac{2}{cosθ}

=2secθ

Problem: 10

Prove that : \frac{sinθ-2sin³θ}{2cos³θ-cosθ}=tanθ

Solution:

LHS = \frac{sinθ-2sin³θ}{2cos³θ-cosθ}

=\frac{sinθ(1-2sin²θ)}{cosθ(2cos²θ-1)}

=\frac{sinθ(1-2sin²θ)}{cosθ(2cos²θ-1)}

= tanθ.\frac{[1-2(1-cos²θ)]}{(2cos²θ-1)}

= tanθ.\frac{(2cos²θ-1)}{(2cos²θ-1)}

=tanθ = RHS

Hence proved

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