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### Practice Set 6.1 Trigonometry Class 10th Mathematics Part 2 MHB Solution

Practice Set 6.1

1. If sintegrate heta = 7/25 find the values of cosθ and tanθ.
2. sin^2theta /costheta +costheta = sectheta Prove that:
3. If tantheta = 3/4 find the values of secθ and cosθ.
4. If cottheta = 40/9 find the values of cosecθ and sinθ.
5. If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
6. If tanθ = 1 then, find the values of sintegrate heta +costheta /sectheta +cosectheta…
7. cos^2theta (1+tan^2theta) = 1 Prove that:
8. root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove that:…
9. (sec θ - cos θ) (cot θ + tan θ) = tan θ sec θ Prove that:
10. cot θ + tan θ = cosec θ sec θ Prove that:
11. 1/sectheta -tantheta = sectheta +tantheta Prove that:
12. sin^4 θ - cos^4 θ = 1 - 2cos^2 θ Prove that:
13. sectheta +tantheta = costheta /1-sintegrate heta Prove that:
14. If tantheta + 1/tantheta = 2 then show that tan^2theta + 1/tan^2theta = 2 Prove that:…
15. tana/(1+tan^2a)^2 + cota/(1+cot^2a)^2 = sinacosa Prove that:
16. sec^4 A (1- sin^4 A) - 2tan^2 A = 1 Prove that:
17. tantheta /sectheta -1 = tantheta +sectheta +1/tantheta +sectheta -1 Prove that:…

###### Practice Set 6.1

We know that,

sin2θ + cos2θ = 1     Also,  Question 2.

Prove that: Taking LHS   [As, sin2θ + cos2θ = 1]

= sec θ = RHS

Proved !

Question 3.

If find the values of secθ and cosθ.

We know that,

sec2θ= 1 + tan2θ    Also,  Question 4.

If find the values of cosecθ and sinθ.

We know that,

cosec2θ = 1 + cot2θ   ⇒ Also, ∴ SinΘ= Question 5.

If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.

5secθ - 12cosecθ = 0

⇒ 5secθ = 12cosecθ   As we have, Also, We know that,

sec2θ= 1 + tan2θ    Also,  Now, again using    Question 6.

If tanθ = 1 then, find the values of Given,

tan θ = 1

⇒ θ = 45° [as tan 45° = 1]

Also,   = sinθ cosθ

= sin 45° cos 45° Question 7.

Prove that: Taking LHS

cos2θ(1 + tan2θ)

= cos2θ sec2θ [As, sec2θ = 1 + tan2θ] = 1

= RHS

Proved !

Question 8.

Prove that: Taking LHS    [(a + b)(a - b) = a2 - b2 ] [As, sin2θ + cos2θ = 1]   = RHS

Proved !

Question 9.

Prove that:

(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ

Taking LHS

(secθ - cosθ)(cotθ + tanθ)

= secθcotθ + secθtanθ - cosθcotθ - cosθtanθ  Taking LCM of first three terms,   [As, sin2θ + cos2θ = 1]

= tanθsecθ

= RHS

Proved !

Question 10.

Prove that:

cot θ + tan θ = cosec θ sec θ

Taking LHS, and putting and = cotθ + tanθ  [As, sin2θ + cos2θ = 1]  = RHS

Proved !

Question 11.

Prove that: Taking LHS   = secθ + tanθ [As, sec2θ = 1 + tan2θ ⇒ sec2θ - tan2θ = 1]

= RHS

Proved !

Question 12.

Prove that:

sin4 θ – cos4 θ = 1 – 2cos2 θ

L.H.S = sin4θ – cos4θ

= (sin2θ – cos2θ)(sin2θ + cos2θ)

= (sin2θ – cos2θ)

= (1 – cos2θ – cos2θ)

= 1- 2cos2θ

Question 13.

Prove that: Taking RHS  (Multiplying both Numerator and Denominator by 1+SinΘ)  [As, sin2θ + cos2θ = 1]  = secθ + tanθ

= LHS

Proved !

Question 14.

Prove that:

If then show that Given, Squaring both side, [(a + b)2 = a2 + b2 + 2ab]  Hence, Proved !

Question 15.

Prove that: Taking RHS    = sinA cos3A + cosA sin3A

= sinAcosA(cos2A + sin2A) [As, sin2θ + cos2θ = 1]

= sinAcosA

= RHS

Proved !

Question 16.

Prove that:

sec4A (1– sin4A) – 2tan2 A = 1

Taking LHS

= sec4A(1 - sin4A) - 2tan2A

= sec4A - sin4A sec4A - 2tan2A = sec4A - tan4A - tan2A - tan2A

= sec4A - tan2A(1 + tan2A) - tan2A

= sec4A - tan2A sec2A - tan2A [As, sec2θ = 1 + tan2θ ]

= sec2A(sec2A - tan2A) - tan2A

= sec2A - tan2A [As, sec2θ = 1 + tan2θ ⇒ sec2θ - tan2θ = 1]

= 1

= RHS

Proved !

Question 17.

Prove that: Taking RHS    [As sec2θ -1 = tan2θ ]  = LHS

Proved.

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