Trigonometry Problems - 1

Trigonometry Problems and solutions.

Level - 1
(P) If secθ+tanθ=3 where θ lies in the first quadrant, then find the value of cosθ
Solution : Given secθ+tanθ=3 ____(a)
sec²θ – tan²θ=1
⇒ (secθ – tanθ)(secθ + tanθ)=1
⇒ (secθ – tanθ)(3) =1 { ∵ Given secθ+tanθ=3}
⇒ (secθ – tanθ)=¹/₃ ____(b)
Add (a) and (b) ⇒ secθ+tanθ+secθ – tanθ=3+¹/₃
⇒ 2secθ =⁽⁹⁺¹⁾/₃
⇒²/_cosθ=¹⁰/₃
⇒cosθ =⁶/₁₀
⇒ cosθ=³/₅
(P) If cosecθ – cotθ =¹/₅, then find the value of sinθ
[Hint : use cosec²θ - cot²θ =1 formula] (Enter your answer in the box try your self)

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Given cosecθ – cotθ= (¹/₅)___(a)
⇒ cosec²θ – cot²θ= 1
⇒(cosecθ – cotθ)(cosecθ + cotθ)= 1
⇒ (¹/₅)(cosecθ + cotθ)=1
⇒ cosecθ + cotθ= 5 ____(b)
⇒ Add (a) and (b)
⇒ cosecθ – cotθ+cosecθ +cotθ= 5 +(¹/₅)
⇒ 2 cosecθ = ²⁶/₅
⇒ cosecθ=¹³/₅
⇒ sinθ = ⁵/₁₃
sinθ= 0.38 (approx)

(P) If a=ccosθ+dsinθ and b=csinθ–dcosθ such that a^m+b^m=c^p+d^q, where m,n,p,q ∈N then find the value of m+n+p+q+42.
Solution :
Given a=ccosθ+dsinθ_____(1)
b=csinθ–dcosθ_______(2)
Squaring and adding (1) and (2)
a²+b²=a=(cosθ+dsinθ)²+(csinθ –dcosθ)²
⇒a²+b²=c²cos²θ+d²sin²θ+2cdcosθsinθ+c²sin²θ+d²cos²θ –2cdsinθcosθ
⇒ a²+b²= c²(cos²θ+sin²θ)+d²(sin²θ+cos²θ)
⇒ a²+b²=c²+d²
Comparing above equation with given equation
a^m+b^m=c^p+d^q
∴ m=2, n=2, p=2, q=2
∴ m+n+p+q+42=2+2+2+2+42=50.
(P) 3sinθ+4cosθ =5, then find the value of 3cosθ – 4sinθ
Enter your answer (try your self)
[Hint : Let 3cosθ – 4sinθ= x after eliminate θ from two equations ]

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Given 3sinθ+4cosθ =5____(1)
Let x=3cosθ –4sinθ_____(2)
Squaring and adding both (1) and (2)
⇒x²+5²=9cos²θ+16sin²θ –24sinθcosθ+9sin²θ+16cos²θ+24sinθcosθ
⇒x²+25=9(cos²θ+sin²θ)+16(sin²θ+cos²θ)
⇒x²+25=25
⇒ x²=0
⇒ x=0
⇒ 3cosθ – 4sinθ=0

(P) cos²θ (1+tan²θ)=

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answer is 1

(P) Prove that cosec²θ+sec²θ = cosec²θsec²θ
Try your self

Check Proof

LHS/-
cosec²θ+sec²θ
=(¹/_sin²θ)+(¹/_cos²θ)
=[^{(cos²θ+sin²θ)}/_{(sin²θ cos²θ)}]
=[¹/_{(sin²θ cos²θ)} ]
=cosec²θsec²θ

(P) Prove that √ sec²θ + cosec²θ=tanθ+cotθ
Proof: LHS/- √ sec²θ + cosec²θ
= √ 1+ tan²θ +1+ cot²θ
{∵ sec² θ –tan² θ=1, cosec² –cot² θ =1}
= √ tan²θ +2tanθcotθ+cot²θ
{∵tanθcotθ=1}
= √ (tanθ +cotθ)²
= tanθ+cotθ
Trigonometry Formula :
✦ sin(90°–θ) = cosθ ; cos(90°–θ) = sinθ ;
(Or) sinA=cosB, if A+B=90°
✦ tan(90°–θ) = cotθ ; cot(90°–θ) = tanθ ;
(Or) tanA=cotB, if A+B =90°
✦ cosec(90°–θ) = secθ ; sec(90°–θ) = cosecθ ;
(Or) cosecA=secB, if A+B=90°
(P) Find

cos37°

sin53°

Answer is 1, { ∵ 37°+53°=90°, sinA=cosB, if A+B=90°}
Try these.
(P)

sin41°

cos49°

Check

answer is 1

(P)

tan54°

cot36°

Check

answer is 1

(P)

cosec32°

sec58°

Check

answer is 1