Practice Set 6.1
- If sintegrate heta = 7/25 find the values of cosθ and tanθ.
- sin^2theta /costheta +costheta = sectheta Prove that:
- If tantheta = 3/4 find the values of secθ and cosθ.
- If cottheta = 40/9 find the values of cosecθ and sinθ.
- If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
- If tanθ = 1 then, find the values of sintegrate heta +costheta /sectheta +cosectheta…
- cos^2theta (1+tan^2theta) = 1 Prove that:
- root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove that:…
- (sec θ - cos θ) (cot θ + tan θ) = tan θ sec θ Prove that:
- cot θ + tan θ = cosec θ sec θ Prove that:
- 1/sectheta -tantheta = sectheta +tantheta Prove that:
- sin^4 θ - cos^4 θ = 1 - 2cos^2 θ Prove that:
- sectheta +tantheta = costheta /1-sintegrate heta Prove that:
- If tantheta + 1/tantheta = 2 then show that tan^2theta + 1/tan^2theta = 2 Prove that:…
- tana/(1+tan^2a)^2 + cota/(1+cot^2a)^2 = sinacosa Prove that:
- sec^4 A (1- sin^4 A) - 2tan^2 A = 1 Prove that:
- tantheta /sectheta -1 = tantheta +sectheta +1/tantheta +sectheta -1 Prove that:…
Practice Set 6.1
If 
find the values of cosθ and tanθ.
Answer:
We know that,
sin2θ + cos2θ = 1
Also,
Question 2.
Prove that:
Answer:
Taking LHS
[As, sin2θ + cos2θ = 1]
= sec θ 
= RHS
Proved !
Question 3.
If 
find the values of secθ and cosθ.
Answer:
We know that,
sec2θ= 1 + tan2θ
Also,
Question 4.
If 
find the values of cosecθ and sinθ.
Answer:
We know that,
cosec2θ = 1 + cot2θ
⇒ 
Also,
∴ SinΘ= 
Question 5.
If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
Answer:
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 
Answer:
Given,
tan θ = 1
⇒ θ = 45° [as tan 45° = 1]
Also,
= sinθ cosθ
= sin 45° cos 45°
Question 7.
Prove that:
Answer:
Taking LHS
cos2θ(1 + tan2θ)
= cos2θ sec2θ [As, sec2θ = 1 + tan2θ]
= 1
= RHS
Proved !
Question 8.
Prove that:
Answer:
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
Answer:
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
Answer:
Taking LHS, and putting 
and 
= cotθ + tanθ
[As, sin2θ + cos2θ = 1]
= RHS
Proved !
Question 11.
Prove that:
Answer:
Taking LHS
= secθ + tanθ [As, sec2θ = 1 + tan2θ ⇒ sec2θ - tan2θ = 1]
= RHS
Proved !
Question 12.
Prove that:
sin4 θ – cos4 θ = 1 – 2cos2 θ
Answer:
L.H.S = sin4θ – cos4θ
= (sin2θ – cos2θ)(sin2θ + cos2θ)
= (sin2θ – cos2θ)
= (1 – cos2θ – cos2θ)
= 1- 2cos2θ
Question 13.
Prove that:
Answer:
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 
Answer:
Given,
Squaring both side,
[(a + b)2 = a2 + b2 + 2ab]
Hence, Proved !
Question 15.
Prove that:
Answer:
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
Answer:
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:
Answer:
Taking RHS
[As sec2θ -1 = tan2θ ]
= LHS
Proved.
