Class 10 Maths Chapter 8 Important Question Answers - Introduction to Trigonometry

Short Answer Type Questions- I

Q1: In ΔABC , Right Angled at B, AB = 24cm, BC = 7cm.
Determine the Following Equations:    
(i) sinA, cosA
Ans: Let us draw a right-angled triangle ABC, right angled at B.

Using Pythagoras theorem, find AC.
AC² = AB²+BC²
= (24)²+(7)²
= 576+49
= 625
∴ AC = 25cm
Then,

(ii) sinC, cosC
Ans: Let us draw a right-angled triangle ABC, right angled at B.

Using Pythagoras theorem, find AC.
AC² = AB²+BC²
= (24)²+(7)²
= 576+49
= 625
∴ AC = 25cm
Then,


Q2: In Adjoining Figure, Find the Value of tanP − cotR.

Ans: Using Pythagoras theorem,
PR² = PQ²+QR²
(13)² = (12)²+QR²
QR² = 169−144⇒25
∴ QR = 5cm
Then find tanP − cotR,
First find the value of tanP.    

Now find the value of cotR
We know that, tanR= 1/cotR
For that we need to first find the value of tanR

Then,
 

Q3: If sinA = 3/4, Calculate the Value of cosA and tanA.  

Ans: Given that the triangle ABC in which ∠B = 90º
Let us take BC = 3k and AC = 4k
Then using Pythagoras theorem,

Calculate the value of cosA

And calculate the value of tanA


Q4: Given 15cotA = 8, Find the Values of sinA and secA.
Ans: Given: 15cotA = 8
Let us assume a triangle ABC in which ∠B = 90º.
Then, 15cotA = 8
⇒ cotA = 8/15
Since cotA =  
Let us draw the triangle.

Now, AB = 8k and BC = 15k.
Using Pythagoras theorem, find the value of AC.

Now, find the values of sinA and secA.


Q5: If ∠A and ∠B are Acute Angles Such That cosA = cosB, then show that ∠A = ∠B

Ans: Given: cosA = cosB
In right triangle ABC,

Now, equate equation (1) and (2).

Therefore, Angles opposite to equal sides are equal.
Hence proved.

Q6: State Whether the Following are True or False. Justify Your Answer.
(i) The Value of tanA is Always Less than 1.
Ans: False because sides of a right triangle may have any length, so tanA may have any value. For example,

(ii) secA= 12/5 for Some Value of Angle A.
Ans: True as secA is always greater than 1. For example,  As hypotenuse will be the largest side. So, it is true.
(iii) cosA  is the Abbreviation Used for the Cosecant of Angle A.
Ans: False as cosA is the abbreviation of cosineA. Because cosA means cosine of angle A and cosecA means cosecant of angle A.
(iv) cotA  is the Product of cotand A.
Ans: False as cotA is not the product of cot and A. cot without A doesn’t have meaning.
(v) sinθ = 4/3 for Some Angle θ.
Ans: False as sinθ cannot be greater than 1. For example, sinθ = Since the hypotenuse is the largest side. So, sinθ will be less than 1.

Q7: Evaluate the Following Equations:
(i)
Ans: Given:
We know that, sin(90°−θ) = cosθ
Then,

(ii)
Ans: Given:
We know that, tan(90°−θ) = cotθ
Then,
 
(iii) cos48°−sin42°
Ans: Given: cos48° − sin42°
We know that, cos(90°−θ) = sinθ
Then,
⇒ cos(90°−42°) − sin42°
⇒ sin42°− sin42° ⇒ 0
∴ cos48° − sin42° = 0
(iv) cosec31°−sec59°
Ans: Given: cosec31°−sec59°
We know that, cosec(90°−θ) = secθ
Then,
⇒ cosec((90°−59°) − sec59°
⇒ sec59°−sec59°⇒0
∴ cosec31° − sec59° = 0

Q8: Show that the Following Equations:
(i) tan48° tan23° tan42° tan67°  = 1
Ans: Given: tan48°tan23°tan42°tan67°=1
We know that, tan(90°−θ)=cotθ.
Now let us take left-hand side,
tan48°tan23°tan42°tan67°
⇒ tan(90°−42°)tan(90°−67°)tan42°tan67°
⇒ cot42°cot67°tan42°tan67°

⇒ 1 is equal to R.H.S
∴tan48°tan23°tan42°tan67° = 1
Hence proved.
(ii) cos38°cos52°−sin38°sin52° = 0
Ans: Given: cos38°cos52°−sin38°sin52° = 0
We know that, cos(90°−θ) = sinθ
Now let us take left-hand side,
cos38°cos52°−sin38°sin52°
⇒ cos(90°−52°)cos(90°−38°)−sin38°sin52°
⇒ sin52°sin38°−sin38°sin52°
⇒ 0 is equal to R.H.S
∴ cos38°cos52°−sin38°sin52° = 0
Hence proved.

Q9: If  tan2A = cot(A−18°) where 2A is an Acute Angle, Find the Value of A.
Ans: Given: tan2A = cot(A−18°)
We know that, cot(90°−θ) = tanθ
Then,
⇒ cot(90°−2A) = cot(A−18°)
Now equalise the angles,
90°−2A = A−18°
−2A−A = −18°−90°
−3A = −108°
A = 108°/3
∴ A = 36°

Q10: If tanA = cotB, then Prove That A+B=90°.
Ans: Given: tanA = cotB
We know that, cot(90°−θ) = tanθ
Then,
cot(90^∘−A) = cotB
Now equalise the angles,
90°−A = B
⇒ A+B = 90∘
∴ A+B = 90°
Hence proved.

Q11: If sec4A = cosec(A−20°), Where 4A is an Acute Angle, Then Find the Value of A.
Ans: Given: sec4A = cosec(A−20°)
We know that, cosec(90°−θ) = secθ
Then,
⇒ cosec(90°−4A) = cosec(A−20°)
Now equalise the angles,
90°−4A = A−20°
−4A−A = −20°−90°
−5A = −110°
A = 110°5
∴ A = 22°
Hence proved.

Q12: If A, B and C are Interior Angles of a ΔABC, then Show That sin
Ans: Given: A, B and C are interior angles of a ΔABC.
We know that, A+B+C = 180°.
Let us consider,

Multiply sin on both sides,

Hence proved.

Q13: Express sin67° + cos75° in terms of trigonometric ratios of angles between  0° and 45°.
Ans: Given: sin67° + cos75°
We know that, sin(90°−θ) = cosθ
and cos(90°−θ) = sinθ.
Then,
sin67° + cos75° = sin(90°−23°) + cos(90°−15°)
= cos23° + sin15°
∴ cos23° + sin15° is the required value.

Q14: Express the Trigonometric Ratios sinA,secA and tanA in Terms of cotA.
Ans: Find the value of sinA in terms of cotA.
By using identity cosec²A−cot²A = 1.
Then, use cosecA= 1/sinA.

Now find the value for secA in terms of cotA.
Using identity sec²A−tan²A = 1
⇒ sec²A = 1+tan²A

And find the value for tanA in terms of cot A
By trigonometric ratio property, tanA= 1/cot A
Hence, tanA = 1/cot A
Therefore, sinA, secA and tanA are founded in terms of cotA.

Q15: Write the Other Trigonometric Ratios of A in Terms of secA.
Ans: Find the value of sinA in terms of secA
By using identity, sin²A+cos²A = 1

Now find the value for cosA in terms of secA,
By trigonometric ratio property, cosA = 1/sec A

Find the value for tanA in terms of secA,
By using identity, sec²A−tan²A = 1
⇒ tan²A = sec²A−1

Find the value for cosecA in terms of secA
By trigonometric ratio property, cosecA = 1/sinA

Substitute the value of sinA =

Finally, find the value for cotA in terms of secA
By trigonometric ratio property, cotA = 1/tan A

Substitute the value of tanA =
 

Q16: Evaluate the Following Equations:  
(i)
Ans: Given:
We know that, sin(90°−θ) = cosθ,cos(90°−θ) = sinθ and  sin²θ+cos²θ = 1
Then,

(ii) sin25°cos65°+cos25°sin65°
Ans: Given: sin25°cos65°+cos25°sin65°
We know that, sin(A+B) = sinAcosB + cosAsinB
Then,
⇒ sin(25°+65°)
⇒ sin90°
⇒ 1
∴ sin25°cos65° + cos25°sin65° = 1

Q17: Show that Any Positive Odd Integer Is of the Form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.  
Ans: Let a be any positive integer and b= 6.
Then, by Euclid’s algorithm, a=6q+r for some integer q ≥ 0, and r = 0,1,2,3,4,5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q+ 1 or 6q+ 2or 6q +3or 6q+ 4or 6q+ 5
Also, 6q+1 = 2×3q+1 = 2k1+1, where k₁ is a positive integer 6q+3 = (6q+2)+1 = 2(3q+1)+1 = 2k₂+ 1,
Where k₂ is an integer 6q+5 = (6q+4)+1 = 2(3q+2)+1 = 2k₃+1, where k₃ is an integer
Clearly, 6q+1, 6q+3, 6q+5 are of the form 2k+ 1, where k an integer is.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q+1, or 6q+3, or 6q+5.

Q18: An Army Contingent of 616 Members are to March Behind an Army Band of 32 Members in a Parade. The Two Groups Are to March in the Same Number of Columns. What Is the Maximum Number of Columns in Which They Can March?
Ans:  We have to find the HCF(616, 32) to find the maximum number of columns in which they can march.
To find the HCF, we can use Euclid’s algorithm.
616 = 32×19+8
⇒ 32 = 8×4+0
Hence,  HCF(616, 32) is 8.
Therefore, they can march in 8 columns each.

Q19: Use Euclid’s Division Lemma to Show That the Square of Any Positive Integer Is Either of Form 3m  or 3m+1 for some integer m.
Hint:Let\[x\]beanypositiveintegerthenitisoftheform\[3q,3q+1\]or\[3q+2 \]. Nowsquareeachoftheseandshowthattheycanberewrittenintheform\[3m\]or\[3m+1\].
Ans: Let a be any positive integer and b=3.
Then a = 3q+r for some integer q ≥ 0
And r = 0,1,2 because 0 ≤ r < 3
Therefore, a = 3q or 3q+1 or 3q+2
Or,
⇒ a² = (3q)² or (3q+1)² or (3q+2)²
⇒ a² = (9q)² or 9q²+6q+1 or 9q²+12q+4
⇒ a² = 3×(3q)² or 3(3q²+2q)+1 or 3(3q²+4q+1)+1
⇒ a = 3k₁ or 3k₂+1 or 3k₃+1
Where k₁, k₂, k₃ are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.

Short Answer Type Questions- II

Q20: Given secθ= 13/12, Calculate the Values for All Other Trigonometric Ratios.

Ans: Given:  secθ = 13/12
Let us consider a triangle ABC in which ∠A = θ and ∠B = 90º
Let AB = 12k and AC = 13k
Then, find the value of BC

Since, secθ = 13/12
Similarly,


Q21: If cotθ= 7/8, then Evaluate the Followings Equations:
(i)
Ans: Given: cotθ= 7/8
Let us consider a triangle ABC in which ∠A = θ and ∠B = 90°
Then, AB = 7k and BC = 8k
Using Pythagoras theorem, find AC


Now find the value of trigonometric ratios.

Then,

We know that, cos²θ+sin²θ = 1
Then,

(ii) cot²θ
Ans: Given: cot²θ
We know that, cotθ= cosθ/sinθ
Then,

∴ cot²θ = 49/64
Hence,  and cot²θ are same.

Q22: If 3cotA = 4, then show that

Ans: Given: 3cotA=4
Let us consider a triangle ABC in which  ∠B = 90º
Then, 3cotA = 4 ⇒ cotA = 4/3
Let AB = 4k and BC = 3k
Using Pythagoras theorem, find AC

Now find the value of trigonometric ratios.

To prove:
Let us take left-hand side

Substitute the value of tanA.

⇒ 7/25

Then,
R.H.S =cos²A−sin²A

⇒ 7/25
∴ cos²A−sin²A = 7/25.
It shows that L.H.S = R.H.S

Hence proved.

Q23: In ΔABC Right Angles at B, if A =  then Find Value of the Following Equations:
(i) sinA cosC + cosA sinC

Ans: Let us consider a triangle ABCin which  ∠B = 90°
Let BC = k and AB = 3–√k
Then, using Pythagoras theorem find AC

Now find the value of trigonometric ratios.

For ∠C, adjacent = BC, opposite = AB, and hypotenuse = AC

Now find the values of the following equations,
sinAcosC + cosAsinC

(ii) cosAcosC−sinAsinC
Ans: 
∴ cosAcosC − sinAsinC = 0

Q24: In ΔPQR, right angled at Q, PR + QR = 25cm and PQ = 5cm. Determine the Values of  sinP, cosP  and tanP.

Ans: Given: In ΔPQR, right angled at Q
And PR+QR = 25cm, PQ = 5cm
Let us take QR = x cm and PR = (25−x)cm
By using Pythagoras theorem, find the value of x.
RP² = RQ²+QP²
⇒ (25−x)²=(x)²+(5)² ⇒ 625 − 50x + x² = x²+25
⇒ −50x =−600 ⇒ x =12
Hence, RQ=12cmand RP=25−12=13cm
Now, find the values of sinP, cosP and tanP.


Q25: If tan(A+B) =√3 and tan(A−B) = 0°<A + B ≤ 90° ; A>B. Find A and B.
Ans: Given: tan(A+B) =√3 and tan(A−B)= 1/√3.
We know that, tan60° =√3 and tan 30° = 1/√3.  
Then,
tan(A+B) = tan60°
⇒ A+B = 60°          …… (1)
tan(A−B) = tan30°
⇒ A−B = 30°      …… (2)
Adding equation (1) and (2). We get,
A+B+A−B = 60°+30° ⇒ 2A=90° ⇒ A = 45°
∴A = 45°
Put A = 45° in equation (1).
A+B = 60°
⇒ 45°+B = 60° ⇒ B= 60°−45° ⇒ B=15°
∴ B = 15°
Hence, A = 45° and B = 15°.

Q26: If xcosθysinθ = a, xsinθ+ycosθ = b, Prove that x²+y² = a²+b².
Ans: Given:
xcosθysinθ = a      …… (1)
xsinθ+ycosθ = b        …… (2)
Squaring and adding the equation (1) and (2) on both sides.
x²cos²θ+y²sin²θ−2xycosθsinθ+x²sin²θ+y²cos²θ+2xycosθsinθ = a²+b²
⇒ x²(cos²θ+sin²θ)+y²(sin²θ+cos²θ) = a²+b²
⇒ x²+y² = a²+b²
∴ x²+y² = a²+b²
Hence proved.

Q27: Prove that sec²θ+cosec²θ Can Never Be Less Than 2.
Ans: Given: sec²θ+cosec²θ
We know that, sec²θ = 1+tan²θ and cosec²θ = 1+cot²θ.
⇒ sec²θ+cosec²θ = 1+tan²θ+1+cot²θ
⇒ sec²θ+cosec²θ = 2+tan²θ+cot²θ
Therefore, sec²θ+cosec²θ can never be less than 2.
Hence proved.

Q28: If sinφ= 1/2, show that 3cosφ−4cos³φ = 0
Ans: Given: sinφ = 1/2
We know that sin30° = 12.
While comparing the angles of sin, we get
⇒ φ = 30°
Substitute φ = 30° to get
3cosφ−4cos³φ = 3cos(30°)−4cos³(30°)

Therefore, 3cosφ−4cos³φ = 0.
Hence proved.

Q29: If 7sin²φ+3cos²φ = 4, then Show that tanφ = 1/√3.
Ans: Given: 7sin²φ+3cos²φ = 4
We know that, sin²φ+cos²φ = 1 and tanθ =
Then, 7sin²φ+3cos²φ = 4(sin²φ+cos²φ)
⇒ 7sin²φ−4sin²φ = 4cos²φ−3cos²φ
⇒ 3sin²φ = cos²φ

Hence proved.

Q30: If cosφ+sinφ = √2cosφ, Prove that cosφ−sinφ = √2sinφ.
Ans: Given: cosφ+sinφ = √2cosφ
Squaring on both sides, we get
⇒ (cosφ+sinφ)² = 2cos²φ
⇒ cos²φ+sin²φ+2cosφsinφ = 2cos²φ
⇒ sin²φ = 2cos²φ−cos²φ−2cosφsinφ
⇒ sin²φ = cos²φ−2cosφsin
Add sin²φ on both sides
⇒ 2sin²φ = cos²φ−2cosφsinφ+sin²φ
⇒ 2sin²φ = (cosφ−sinφ)²
∴ cosφ−sinφ = √2 sinφ
Hence proved.

Q31: If tanA+sinA = m and tanA−sinA = n, then Show that m²−n² = 4√mn.
Ans: Given:
tanA + sinA = m        …… (1)
tanA − sinA = n         …… (2)
Now to prove m²−n²= 4√mn.
Take left-hand side

Now take right-hand side

Hence, 4√mn = 4tanAsinA
∴ m²−n² = 4√mn
Hence proved.

Q32: If secA = x + (1/4x), then prove that secA+tanA = 2x or (1/2x).
Ans: Given: secA = x+ (1/4x)
Squaring on both sides.

We know that, sec²A = 1+tan²A


Taking square root on both sides,

Now, find secA + tanA
If tanA = x − (1/4x)means 

Hence proved.

Q32: If A, B are Acute Angles and sinA = cosB, then Find the Value of A+B.
Ans: Given: sinA = cosB
We know that sinA = cos(90°−A)
While comparing the values to get
cosB = cos(90°−A)
⇒ B = 90°−A ⇒ A+B = 90°
∴ A+B = 90°.

Q33: Evaluate the Following Questions:
(i) Solve for ϕ, if tan5ϕ = 1.
Ans: Given: tan5ϕ = 1
We know that, tan⁻¹(1) = 45°
5ϕ = tan⁻¹(1) ⇒ 45°
5ϕ = 45°
ϕ = 45°/5 ⇒ 9°
∵ϕ = 9°
(ii) Solve for φ, if
Ans:


sinφ = sin30° ⇒ φ = 30°
∴ φ = 30°.

Q34: If  show that (m²+n²)cos²β = n².
Ans: Given: … (1)
 … (2)
Squaring equation (1) and (2). We get,

Now to prove (m²+n²)cos²β = n²,
Take left-hand side,


= n²
∴ (m²+n²)cos²β = n²
Hence proved.

Q35: If 7cosecφ−3cotφ = 7, then prove that 7cotφ−3cosecφ = 3.
Ans: Given: 7cosecφ−3cotφ=7
Then prove that, 7cotφ−3cosecφ=3
7cosecφ−3cotφ=7
Squaring on both sides, we get
49cosec²φ+9cot²φ−42cosecφcotφ=49
We know that, cosec²φ=1+cot²φ and cot²φ=cosec²φ−1.
49(cot²φ+1)+9(cosec²φ−1)−42cosecφcotφ=49
49cot²φ+49+9cosec²φ−9−2(3cosecφ⋅7cotφ)=49
(7cotφ−3cosecφ)² = 49−49+9
(7cotφ−3cosecφ)² = 9
Take square root on both sides, we get
∴ 7cotφ−3cosecφ = 3
Hence proved.

Q36: Prove that 2(sin⁶φ+cos⁶φ) 3(sin⁴φ+cos⁴φ)+1 = 0.
Ans: Given: 2(sin⁶φ+cos⁶φ) 3(sin⁴φ+cos⁴φ)+1=0
Let us take left-hand side,
2(sin⁶φ+cos⁶φ) 3(sin⁴φ+cos⁴φ)+1=2((sin²φ)³+(cos²φ)³)−3((sin²φ)²+(cos²φ)²)+1
=2[(sin²φ+cos²φ)³−3sin²φcos²φ(sin²φ+cos²φ)]−3[(sin²φ+cos²φ)²−2sin²φcos²φ]+1
=2[1−3sin²φcos²φ]−3[1−2sin²φcos²φ]+1
=2−6sin²φcos²φ−3+6sin²φcos²φ+1
=−1+1
= 0
Therefore, 2(sin⁶φ+cos⁶φ) 3(sin⁴φ+cos⁴φ)+1 = 0.
Hence proved.

Q37: If tanθ = 5/6 and θ = ϕ = 90°. What is the value of cotϕ.
Ans: Given: tanθ = 5/6 and θ = ϕ = 90°
We know that, tanθ = 1/cotθ.
cotϕ = 1/tanϕ
= 1/(5/6)
= 6/5
∴cotϕ= 6/5.

Q38: What is the Value of tanφ in terms of sinφ ?
Ans: Given: tanφ
We know that, tanφ =  and cos²φ+sin²φ = 1
tanφ = sinφ/cosφ
 

Q39: If secφ+tanφ = 4, Find the Value of sinφ, cosφ.
Ans: Given: secφ+tanφ = 4

Squaring on both sides.
(1+sinφ)² = (4cosφ)²
1+2sinφ+sin²φ = 16cos²φ
1+2sinφ+sin²φ = 16(1−sin²φ)
1+2sinφ+sin²φ = 16−16sin²φ
17sin²φ+2sinφ−15 = 0
17sin²φ+17sinφ−15sinφ−15 = 0
17sinφ(sinφ+1)−15(sinφ+1) = 0
(sinφ+1)(17sinφ−15) = 0
If sinφ+1 = 0
Hence, sinφ = −1 is not possible.
Then, 17sinφ−15 = 0
∴ sinφ = 15/17
Now find cosφ

Substitute the value of sinφ
 1 + (15/17) = 4cosφ
32/17 = 4cosφ
⇒ cosφ =
∴ cosφ = 8/17.

Long Answer Type Questions

Q1: Express the Trigonometric Ratios sinA, secA and tanA in Terms of cotA.
Ans: Find the value for sinA in terms of cotA
By using identity cosec²A−cot²A = 1
Then,
⇒ cosec²A = 1+cot²A

Express the value of secA in terms of cotA
By using identity sec2A−tan²A = 1
Then,

Express the value of tanA in terms of cotA
We know that, tanA = 1/cotA
∴ tanA =1/cotA.

Q2: Write the Other Trigonometric Ratios of A in Terms of secA.
Ans: Express the value of sinA in terms of secA
By using identity, sin²A+cos²A = 1
⇒ sin²A = 1−cos²A


Express the value of cosA in terms of secA
We know that, cosA = 1/secA
∴ cosA = 1/secA
Express the value of tanA in terms of secA
By using identity sec²A−tan²A = 1
Then,
⇒ tan²A = sec²A−1

Express the value of cosecA in terms of secA
We know that, cosecA= 1sinA
Then, Substitute the value of sinA

Express the value of cotA in terms of secA
We know that,
Substitute the value of tanA


Q3: Evaluate the following equations:
(i)
Ans: Given:
We know that, sin(90°−θ) = cosθ, cos(90°−θ) = sinθ and sin²θ+cos²θ = 1
Then,

(ii) sin25°cos65°+cos25°sin65°
Ans: Given: sin25°cos65°+cos25°sin65°
We know that, sin(90°−θ) = cosθ, cos(90°−θ) = sinθ and sin²θ+cos²θ = 1
⇒ sin25°cos(90°−25°) + cos25°sin(90°−25°)
⇒ sin25°.sin25°+cos25°.cos25°
⇒ sin225°+cos225° = 1
∴ sin25°cos65° + cos25°sin65° = 1

Q4: Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:
(i) (cosecθ−cotθ)²=
Ans: Given: (cosecθ−cotθ)²=
We know that, (a−b)² = a²+b²−2ab, cosecθ =
Then, let us take left-hand side
⇒ (cosecθ−cotθ)² = cosec²θ + cot²θ − 2cosecθcotθ

Hence proved.
(ii)
Ans: Given:
We know that, sin²θ+cos²θ = 1
Then, let us take left-hand side


Hence proved.
(iii)
Ans: Given:
We know that, a³−b³= (a−b)(a²+b²+ab) and sin²θ+cos²θ = 1
Then, let us take L.H.S


(iv)
Ans: Given:

Then, let us take L.H.S

Hence proved.
(v)  using the identity cosec²A = 1+cot²A
Ans: Given:
We know that, cosec²A = 1+cot²A
Then, let us take L.H.S
 
Dividing all terms by sinA


Hence proved.

(vi)
Ans: Given:
We know that, 1−sin²θ=cos²θ and (a+b)(a−b) = a²−b²
Then, let us take L.H.S
 
Let us take conjugate of the term. Then,

Hence proved.  
(vii)
Ans: Given:
We know that, 1−sin²θ = cos²θ
Then, let us take L.H.S

Hence proved.

(viii) (sinA+cosecA)²+(cosA+secA)² = 7+tan²A+cot²A
Ans: Given: (sinA+cosecA)²+(cosA+secA)²=7+tan²A+cot²A
We know that, cosec²θ =1+cot²θ and sec²θ = 1+tan²θ
Then, let us take L.H.S


= 5+cosec²A+sec²A
= 5+1+cot²A+1+tan²A
= 7+tan²A+cot²A = R.H.S
∴ (sinA+cosecA)²+(cosA+secA)²=7+tan²A+cot²A
Hence proved.
(ix) (cosecA−sinA)(secA−cosA)=
Ans: Given: (cosecA−sinA)(secA−cosA) =
We know that, sin²θ+cos²θ = 1
Then, let us take L.H.S
(cosecA−sinA)(secA−cosA) =

Dividing all the terms by sinA.cosA

∴ (cosecA−sinA)(secA−cosA) =
Hence proved.
(x)    
Ans: Given:

We know that, 1+tan²θ = sec²θ and 1+cot²θ = cosec²A
Then, let us take L.H.S

Now, prove the Middle side


= (−tanA)²
= tan²A = R.H.S

Hence proved.

Q6: Use Euclid’s Division Algorithm to Find the HCF of:
(i) 135 and 225
Ans: Given: 135 and 225
We have 225>135,
So, we have to apply the division lemma to 225 and 135 to obtain
225=135×1+90
Here remainder90 ≠ 0, again we are applying the division lemma to 135 and 90 to obtain 135=90×1+45
Again, the remainder45≠0, then apply the division lemma to obtain
90 = 2×45+0
Since now we got remainder as zero. Here, the process get stops.
The divisor at this stage is 45
Therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
Ans: Given: 196 and 38220
We have 38220 > 196,
So, we have to apply the division lemma to 38220 and 196 to obtain
38220 = 196×195+0
Since we get the remainder as zero, the process stops here.
The divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255
Ans: Given: 867 and 255
We have 867 > 255,
So, we have to apply the division lemma to 867 and 255 to obtain
867 = 255×3+102
Here remainder 102 ≠ 0, again apply the division lemma to255 and 102 to obtain
255 = 102×2 51
Again, remainder 51≠0, again apply the division lemma to 102 and 51 to obtain
102 = 51×2+0
Since we get the remainder as zero, the process stops here.
The divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.

Q7: Evaluate the Following Equations:  
i. sin60°cos30°+sin30°cos60°
Ans: Given: sin60°cos30°+sin30°cos60°
We know that,
Then, sin60°cos30°+sin30°cos60°

= 4/4 = 1
∴ sin60°cos30°+sin30°cos60° = 1
(ii) 2tan²45°+cos²30°−sin²60°
Ans: Given: 2tan²45°+cos²30°−sin260°
We know that, tan45° = 1, sin60° = √3/2 and cos30° = √3/2
Then, 2tan²45°+cos²30°−sin²60°

(iii)
Ans: Given:

We know that,



(iv)
Ans: Given:
We know that,  
Then,


(v)
Ans: Given:
 Then,




Q8: Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:
(i) (cosecθ−cotθ)²=
Ans: Given: (cosecθ−cotθ)²=
We know that, (a−b)²= a²+b²−2ab, cosecθ =
Then, let us take left-hand side
⇒(cosecθ−cotθ)²=cosec²θ+cot²θ−2cosecθcotθ

Hence proved.
(ii)
Ans: Given:
We know that, sin²θ+cos²θ =1
Then, let us take left-hand side



Hence proved.
(iii)
Ans: Given:

We know that, a³−b³=(a−b)(a²+b²+ab) and sin²θ+cos²θ=1
Then, let us take L.H.S


Hence proved.
(iv)
Ans: Given:

Then, let us take L.H.S

Hence proved.
(v)  using the identity cosec²A=1+cot²A
Ans: Given:

We know that, cosec²A=1+cot²A
Then, let us take L.H.S

Dividing all terms by sinA


Hence proved.
(vi)
Ans: Given:

We know that, 1−sin²θ=cos²θ and (a+b)(a−b)=a²−b²
Then, let us take L.H.S

Let us take conjugate of the term. Then,

Hence proved.
(vii)
Ans: Given:

We know that, 1−sin²θ=cos²θ
Then, let us take L.H.S

Hence proved.
(viii) (sinA+cosecA)²+(cosA+secA)²=7+tan²A+cot²A
Ans: Given: (sinA+cosecA)²+(cosA+secA)²=7+tan²A+cot²A
We know that, cosec²θ=1+cot²θ and sec²θ=1+tan²θ
Then, let us take L.H.S
(sinA+cosecA)²+(cosA+secA)²=

= 5+cosec²A+sec²A
= 5+1+cot²A+1+tan²A
= 7+tan²A+cot²A = R.H.S
∴ (sinA+cosecA)²+(cosA+secA)²=7+tan²A+cot²A
Hence proved.
(ix) (cosecA−sinA)(secA−cosA) =
Ans: Given: (cosecA−sinA)(secA−cosA)=
We know that, sin²θ+cos²θ = 1
Then, let us take L.H.S
(cosecA−sinA)(secA−cosA)=

Dividing all the terms by sinA.cosA

∴(cosecA−sinA)(secA−cosA) =
Hence proved.
(x)
Ans: Given:

We know that, 1+tan²θ=sec²θ and 1+cot²θ=cosec²A
Then, let us take L.H.S

Now, prove the Middle side


= (−tanA)²
= tan²A = R.H.S

Hence proved.