### Prove the following identities (1 – 13)

### Question 1. sec^{4}θ – sec^{2}θ = tan^{4}θ + tan^{2}θ

**Solution:**

We have

sec

^{4}θ – sec^{2}θ = tan^{4}θ + tan^{2}θTaking LHS

= sec

^{4}θ – sec^{2}θ= sec

^{2}θ(sec^{2}θ – 1)Using sec

^{2}θ = tan^{2}θ + 1, we get= (1 + tan

^{2}θ)tan^{2}θ= tan

^{2}θ + tan^{4}θHence, LHS = RHS (Proved)

### Question 2. sin^{6}θ + cos^{6}θ = 1 – 3sin^{2}θcos^{2}θ

**Solution:**

We have

sin

^{6}θ + cos^{6}θ = 1 – 3sin^{2}θcos^{2}θTaking LHS

= sin

^{6}θ + cos^{6}θ= (sin

^{2}θ)^{3}+ (cos^{2}θ)^{3}Using a

^{3 }+ b^{3 }= (a + b)(a^{2 }+ b^{2 }– ab), we get= (sin

^{2}θ + cos^{2}θ)(sin^{4}θ + cos^{4}θ – sin^{2}θcos^{2}θ)Using a

^{2 }+ b^{2 }= (a + b)^{2 }– 2ab and sin^{2}θ + cos^{2}θ = 1, we get= (1)[(sin

^{2}θ + cos^{2}θ)^{2 }– 2sin^{2}θcos^{2}θ – sin^{2}θcos^{2}θ]= (1)[(1)

^{2 }– 3sin^{2}θcos^{2}θ]= 1 – 3sin

^{2}θcos^{2}θHence, LHS = RHS (Proved)

### Question 3. (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1

**Solution:**

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)

### Question **4. **cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ

**Solution:**

We have

cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ

Taking LHS

=

=

=

=

=

=

=

Hence, LHS = RHS(Proved)

### Question 5.

**Solution:**

We have

Taking LHS

=

Using a

^{2 }– b^{2 }= (a + b)(a – b) and a^{3 }+ b^{3 }= (a + b)(a^{2 }+ b^{2}ab), we get=

=

=

=

= sinA

Hence, LHS = RHS(Proved)

### Question 6.

**Solution:**

We have

Taking LHS

=

Using tanA = sinA/cosA and cotA = cosA/sinA, we get

=

=

=

=

Using a

^{3}– b^{3}= (a – b)(a^{2}+ b^{2}+ ab), we get=

=

=

Using cosecA = 1/sinA and secA = 1/cosA, we get

= secAcosecA + 1

Hence, LHS = RHS(Proved)

### Question 7.

**Solution: **

We have

Taking LHS

=

Using a

^{3 }± b^{3 }= (a ± b)(a^{2 }+ b^{2 }± ab), we get=

Using sin

^{2}θ + cos^{2}θ = 1, we get= 1 – sinAcosA + 1 + sinAcosA

= 2

Hence, LHS = RHS(Proved)

### Question 8. (secAsecB + tanAtanB)^{2 }– (secAtanB + tanAsecB)^{2 }= 1

**Solution:**

We have

(secAsecB + tanAtanB)

^{2}– (secAtanB + tanAsecB)^{2}= 1Taking LHS

= (secAsecB + tanAtanB)

^{2 }– (secAtanB + tanAsecB)^{2}Expanding the above equation using the formula

(a + b)

^{2 }= a^{2 }+ b^{2 }+ 2ab= (secAsecB)

^{2 }+ (tanAtanB)^{2 }+ 2(secAsecB)(tanAtanB) –(secAtanB)

^{2 }– (tanAsecB)^{2 }– 2(secAtanB)(tanAsecB)= sec

^{2}Asec^{2}B + tan^{2}Atan^{2}B – sec^{2}Atan^{2}B – tan^{2}Asec^{2}B= sec

^{2}A(sec^{2}B – tan^{2}B) – tan^{2}A(sec^{2}B – tan^{2}B)= sec

^{2}A – tan^{2}A -(Using sec^{2}θ – tan^{2}θ = 1)= 1

Hence, LHS = RHS(Proved)

### Question 9.

**Solution:**

We have

Taking RHS

=

=

= ×

=

=

=

=

=

=

=

=

=

= ×

=

=

=

=

Hence, RHS = LHS(Proved)

### Question 10.

**Solution:**

We have

Taking LHS

=

Using 1 + tan

^{2}x = sec^{2}x and 1 + cot^{2}x = cosec^{2}x, we get=

=

=

=

=

Using a

^{2}+ b^{2}= (a + b)^{2}– 2ab, we get=

=

=

Hence, LHS = RHS (Proved)

### Question 11.

**Solution:**

We have

Taking LHS

=

By using the formulas cotθ = cosθ/sinθ and tanθ = sinθ/cosθ, we get

=

=

=

Using a

^{3}+b^{3 }= (a + b)(a^{2 }+ b^{2 }– ab), we get=

=

= 1 – (sin

^{2}θ + cos^{2}θ) + sinθcosθ= 1 – 1 + sinθcosθ

= sinθcosθ

Hence, LHS = RHS (Proved)

### Question 12.

**Solution:**

We have

=

Taking LHS

=

=

=

=

=

=

=

=

=

Hence, LHS = RHS(Proved)

### Question 13. (1 + tanαtanβ)^{2 }+ (tanα – tanβ)^{2 }= sec^{2}αsec^{2}β

**Solution:**

We have

(1 + tanαtanβ)

^{2 }+ (tanα – tanβ)^{2 }= sec^2αsec^{2}βTaking LHS

= (1 + tanαtanβ)

^{2 }+ (tanα – tanβ)^{2}= (1 + tan

^{2}αtan^{2}β + 2tanαtanβ) + (tan^{2}α + tan^{2}β – 2tanαtanβ)= 1 + tan

^{2}αtan^{2}β + tan^{2}α + tan^{2}β= (1 + tan

^{2}β) + tan^{2}α(1 + tan^{2}β)= (1 + tan

^{2}β)(1 + tan^{2}α)= sec

^{2}αsec^{2}βHence, LHS = RHS (Proved)

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