Trigonometry Questions and Answers

Trigonometry Questions and Answers

Solved Examples

Question 1: Prove the following identity:

Solution:

LHS = (secθ$\theta$ - coseθ$\theta$)(1 + tanθ$\theta$ + cotθ$\theta$)
= (sinθ−cosθsinθcosθ$\frac{sin\theta -cos\theta }{sin\theta cos\theta }$)(sinθcosθ+1sinθcosθ$\frac{sin\theta cos\theta +1}{sin\theta cos\theta }$)
[Using identities, secθ$\theta$ = 1cosθ$\frac{1}{cos\theta }$, coseθ$\theta$ = 1sinθ$\frac{1}{sin\theta }$, tanθ$\theta$ = sinθcosθ$\frac{sin\theta }{cos\theta }$ and cotθ$\theta$ = cosθsinθ$\frac{cos\theta }{sin\theta }$]
= (sinθ−cosθ)(sinθ cosθ+1)sin2θ cos2θ$\frac{(sin\theta -cos\theta )(sin\theta cos\theta +1)}{si{n}^2\theta co{s}^2\theta }$
= (sinθ−cosθ)(sinθ cosθ+sin2θ+cos2θ)sin2θ cos2θ$\frac{(sin\theta -cos\theta )(sin\theta cos\theta +si{n}^2\theta +co{s}^2\theta )}{si{n}^2\theta co{s}^2\theta }$
[Using, sin2θ+cos2θ$si{n}^2\theta +co{s}^2\theta$ = 1]





= tanθ$\theta$ sec θ$\theta$ - cotθ$\theta$coscθ$\theta$
= RHS

Question 2:Prove the following identity:


Solution:

LHS = 1 + 2sec²A tan²A - sec⁴A - tan⁴A

= 1 - (sec⁴A - 2sec²A tan²A + tan⁴A)





[Using identity, sec²A - tan²A = 1]

= 1 - 1

= 0

= RHS

Question 3: Prove the following identity:

Solution:

LHS = [1−sinAcosAcosA(secA−cosecA)$\frac{1-sinAcosA}{cosA(secA-cosecA)}$][sin2A−cos2Asin3A+cos3A$\frac{si{n}^{2}A-co{s}^{2}A}{si{n}^{3}A+co{s}^{3}A}$]

= [1−sinAcosAcosA(1cosA−1sinA)$\frac{1-sinAcosA}{cosA(\frac{1}{cosA}-\frac{1}{sinA})}$][(sinA+cosA)(sinA−cosA)(sinA+cosA)(sin2A−sinAcosA+cos2A)$\frac{(sinA+cosA)(sinA-cosA)}{(sinA+cosA)(si{n}^{2}A-sinAcosA+co{s}^{2}A)}$]

[Using identities, secθ$\theta$ = 1cosθ$\frac{1}{cos\theta }$, coseθ$\theta$ = 1sinθ$\frac{1}{sin\theta }$, sin²A - cos²A = (sin A + cos A)(sin A - cos A) and sin³A + cos³A = (sin A + cos A)(sin² A - sin A cos A + cos²A)]

= [1−sinAcosAsinA−cosAsinA$\frac{1-sinAcosA}{\frac{sinA-cosA}{sinA}}$][sinA−cosAsin2A−sinAcosA+cos2A$\frac{sinA-cosA}{si{n}^{2}A-sinAcosA+co{s}^{2}A}$] [By cancelling common terms]

= sin A[1−sinAcosAsinA−cosA$\frac{1-sinAcosA}{sinA-cosA}$] * [sinA−cosA1−sinAcosA$\frac{sinA-cosA}{1-sinAcosA}$]

[Using identity sin² A + cos² A = 1]

= sin A (By cancelling common terms)

= RHS

Trigonometry Test Questions

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Lets us solve some more trigonometric examples using their identities:

Solved Examples

Question 1: Prove the following identity:


Solution:

LHS = (secθ$\theta$ - 1)² - (tanθ$\theta$ - sinθ$\theta$)²

= (1cosθ$\frac{1}{cos\theta }$ - 1)² - (sinθcosθ$\frac{sin\theta }{cos\theta }$ - sinθ$\theta$)²

= (1−cosθcosθ$\frac{1-cos\theta }{cos\theta }$)² - sin2θcos2θ$\frac{si{n}^2\theta }{co{s}^2\theta }$(1 - sinθ$\theta$ * cosθsinθ$\frac{cos\theta }{sin\theta }$)²

= (1−cosθcosθ$\frac{1-cos\theta }{cos\theta }$)² - (1 - cosθ$\theta$)² sin2θcos2θ$\frac{si{n}^2\theta }{co{s}^2\theta }$
= (1 - cosθ$\theta$)² [1cos2θ−sin2θcos2θ$\frac{1}{co{s}^2\theta }-\frac{si{n}^2\theta }{co{s}^2\theta }$]

Question 2:Prove the following identity:


Solution:

LHS = tan3θ1+tan2θ$\frac{ta{n}^3\theta }{1+ta{n}^2\theta }$ + cot3θ1+cot2θ$\frac{co{t}^3\theta }{1+co{t}^2\theta }$

= tan3θsec2θ$\frac{ta{n}^3\theta }{se{c}^2\theta }$ + cot3θcose2θ$\frac{co{t}^3\theta }{cos{e}^2\theta }$

Trigonometry Sample Questions

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Solved Examples

Question 1:

Solution:

If tan A + sin A = m .................. (1)
tan A - sin A = n ................... (2)
Step 1:Adding (1) and (2)

Step 2:
Subtracting (2) from (1)

Now


[Using, (a + b)² = a² + b² + 2ab and (a - b)² = a² + b² - 2ab]

Question 2: 
Solution:

= 12√+1$\frac{1}{\sqrt{2}+1}$ * 2√−12√−1$\frac{\sqrt{2}-1}{\sqrt{2}-1}$ cos θ$\theta$

= (2–√$\sqrt{2}$ - 1)cos θ$\theta$

Question 3:If x sin³θ$\theta$ + y cos³θ$\theta$ = sin θ$\theta$ cos θ$\theta$ and x sin θ$\theta$ - y cos θ$\theta$ = 0. Prove that x² + y² = 1.

Solution:

x sin³θ$\theta$ + y cos³θ$\theta$ = sin θ$\theta$ cos θ$\theta$ ..................... (i)

x sin θ$\theta$ - y cos θ$\theta$ = 0 ...................... (ii)



from (i) and (iii)







y = sinθ$\theta$ .................. (iv)

From (iii) and (iv), x = cosθ$\theta$
=> x² + y² = sin²θ$\theta$ + cos²θ$\theta$