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GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

cos 60° cos 30° − sin 60° sin 30° = ?

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જવાબ : cos 60^o cos 30^o − sin 60^o sin 30^o
=( 1/2 × √3/2 - √3/2 × ½ ) = (√3/4 - √3/4)=0

Evaluate:

sin30°cos45°+cot45°sec60°-sin60°tan45°+cos30°sin90°

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જવાબ : sin30°cos45°+cot45°sec60°-sin60°tan45°+cos30°sin90° =(1/2)(1/√2)+1/2-(√3/2)1+(√3/2)1 =√2/2 + ½ - √3/2 + √3/2 =√2+1/2

sin 60° cos 30° + cos 60° sin 30° = ?

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જવાબ : sin 60^o cos 30^o + cos 60^o sin 30^o
=(√3/2 × √3/2 + ½ × ½ ) = (3/4 +1/4 )=4/4=1

cos 45° cos 30° + sin 45° sin 30° = ?

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જવાબ : cos 45^o cos 30^o + sin 45^o sin 30^o = ?
= (1/√2 × √3/2 + 1/√2 × ½ ) = (√3/2√2 + 1/2√2) = (√3 +1/2√2)

Evaluate:

5cos²60°+4sec²30°-tan²45°sin²30°+cos²30°

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જવાબ : 5cos²60°+4sec²30°-tan²45°sin²30°+cos²30°= 5(1/2)² + 4(2√3)² -(1)²(1/2)²+(√3/2)² =(5/4+4×4/3-1)(1/4+3/4) =(5/4+16/3-1)(4/4) =[(15+64-12)/12](4/4) =(67/12)(1) =67/12

2 cos² 60° + 3 sin² 45° − 3 sin² 30° + 2 cos² 90° = ?

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જવાબ : 2 cos² 60^o + 3 sin² 45^o − 3 sin² 30^o + 2 cos² 90^o
=2×(1/2)²+3×(1/√2)²-3×(1/2)² + 2×(0)² =2×1/4 + 3×1/2- 3×1/4+0 =(1/2+3/2-3/4)
=[(2+6-3)/4] =54

cot²30° − 2cos²30° − 3/4sec²45° + 1/4 cosec²30° = ?

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જવાબ : cot² 30^o − 2 cos² 30^o – 3/4sec² 45^o + 1/4cosec²30^o
= (√3)²-2×(√3/2)²-3/4×(√2)²+1/4×(2)² = 3-2×3/4-3/4×2+1/4×4 = 3-3/2-3/2+1 = 4-(3/2+3/2) = 4-3= 1

(sin²30° + 4cot²45° − sec²60°)(cosec²45° sec²30°) = ?

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જવાબ : (sin² 30^o + 4 cot² 45^o − sec² 60^o )(cosec² 45^osec² 30^o)
=[(1/2)²+4×(1)²-(2)²] [(√2)² (2/√3)²]=(1/4+4-4) (2×4/3) =1/4×8/3=2/3

4/cot²30°+1/sin²30°-2cos²45°-sin²0° = ?

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જવાબ : 4cot²30° +1sin²30° -2 cos²45^o -sin²0^o =4(√3)²+1(1/2)² - 2×(1/√2)²-(0)² =4/3+1/4-2×1/2-0 =4/3+4-1 =4/3+3 =(4+9)/3 = 13/3

(1-sin60°)/cos60°= (tan60°-1)/tan60°+1

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જવાબ : LHS=1-sin 60ocos 60o=1-√3212=(2-√32)12=(2-√32)×2=2-√3RHS= tan 60o-1tan 60o+1=√3-1√3+1=√3-1√3+1×√3 -1√3 -1=(√3-1)2(√3)2-12=3+1-2√33-1=4-2√32=2-√3

Hence, LHS = RHS
∴ 1-sin 60ocos 60o=tan 60o-1tan 60o+1

Show that:
(cos30° + sin60°)/(1+sin30°+cos60°) = cos30°

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જવાબ : LHS = (cos 30° + sin 60°)/(1 + sin 30° + cos 60°) = = √32 Also, RHS = cos 30° =√3/2

Hence, LHS = RHS
∴ (cos30° + sin60°)/(1+sin30°+cos60°) = cos30°

Verify:
sin 60° cos 30° − cos 60° sin 30° = sin 30°

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જવાબ : sin 60^o cos 30^o − cos 60^o sin 30^o
=(3√2)×(3√2)−(1/2)×(1/2) =3/4−1/4 =2/4 =1/2 Also, sin 30^o=1/2

Verify:
cos 60° cos 30° + sin 60° sin 30° = cos 30°

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જવાબ : cos 60^o cos 30^o + sin 60^o sin 30^o
=(1/2)×( √3/2)+(√3/2)×(1/2) =√3/4+√3/4 =√3/2 Also, cos 30^o =√3/2
∴ cos 60^o cos 30^o + sin 60^o sin 30^o = cos 30^o

Verify:

2 sin 30° cos 30° = sin 60°

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જવાબ : 2 sin 30^o cos 30^o
=2×1/2×3√2 = 3√2 Also, sin 60^o = 3√2
∴ 2 sin 30^o cos 30^o = sin 60^o

Verify:

2 sin 45° cos 45° = sin 90°

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જવાબ : 2 sin 45^o cos 45^o
=2×1/√2×1/√2 =1
Also, sin 90^o = 1
∴ 2 sin 45^o cos 45^o = sin 90^o

If A = 30°, verify that: tan2A =

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જવાબ : tan 2A = tan 60^o = √3
= = = = √3
∴ tan2A =

If A = 30°, verify that: cos2A =

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જવાબ : A = 30^o
⇒ 2A = 2 × 30^o = 60^o
cos 2A = cos 60^o = 1212 = = = = = ½ ∴ cos 2A=

If A = 30°, verify that: sin2A=

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જવાબ : A = 30^o
⇒ 2A = 2 × 30^o = 60^o
sin 2A = sin 60^o = √3/2
= = = = = ∴ sin 2A=2tan A1+tan2Asin 2A=2tan A1+tan2A

If A = 45°, verify that: cos 2A = 2 cos² A − 1 = 1 − 2 sin² A

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જવાબ : A = 45^o
⇒ 2A = 2 × 45^o = 90^o
cos 2A = cos 90^o = 0
2 cos² A − 1 = 2 cos² 45^o − 1 =  2×(1/√2)² − 1 = 2×1/2 −1 = 1−1 = 0
Now, 1 − 2 sin² A = 1−2×(1/√2)² = 1 − 2×1/2  =1 − 1 = 0
∴ cos 2A =  2 cos² A − 1 = 1 − 2 sin² A

If A = 45°, verify that: sin 2A = 2 sin A cos A

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જવાબ : A = 45^o
⇒ 2A = 2 × 45^o = 90^o
sin 2A = sin 90^o = 1
2 sin A cos A = 2 sin 45^o cos 45^o = 2×1/√2×1/√2 = 2×1/2 = 1
∴ sin 2A = 2 sin A cos A

tan²θ/(1+tan²θ) + cot²θ/(1+cot²θ) = 1

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જવાબ : LHS = tan²θ/(1+tan²θ) + cot²θ/(1+cot²θ) =tan²θ/sec²θ + cot²θ/cosec²θ (∵sec2θ−tan2θ=1 and cosec2θ−cot2θ=1) =sin2θcos2θ1cos2θ + cos2θsin2θ1sin2θ =sin²θ+cos²θ =1 =RHS
Hence, LHS = RHS

Prove that, (1+tan²θ)cotθ/cosec²θ=tanθ

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જવાબ : LHS= (1+tan²θ)cotθ/cosec²θ =sec²θcotθ/cosec²θ = 1cos2θ × cosθsinθ1sin2θ = sin²θ/cosθsinθ =sinθ/cosθ =tanθ
Hence, L.H.S. = R.H.S.

Prove that,1+ tan²θ/(1+secθ) = secθ

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જવાબ : LHS=1+ tan²θ/(1+secθ) =1+ (sec²θ−1)/(secθ+1) =1+ (secθ+1)/(secθ−1)(secθ+1) =1+(secθ−1) =secθ =RHS

Prove that,1+cot2θ/(1+cosecθ)=cosecθ

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જવાબ : LHS=1+cot²θ(1+cosecθ) =1+(cosec²θ−1)(cosecθ+1) (∵cosec²θ-cot²θ=1) =1+ (cosecθ+1)(cosecθ−1)/(cosecθ+1) =1+(cosecθ−1) =cosecθ =RHS

Prove that, sec θ (1 − sin θ) (sec θ + tan θ) = 1

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જવાબ : LHS =secθ(1−sinθ)(secθ+tanθ) =(secθ−secθsinθ)(secθ+tanθ) =(secθ−1cosθ×sinθ)(secθ+tanθ) =(secθ−tanθ)(secθ+tanθ) =sec²θ−tan²θ =1 =RHS

Prove that, 1/(1+sinθ)+1/(1-sinθ)=2sec²θ

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જવાબ : LHS=1/(1+sinθ)+1/(1-sinθ) = [(1-sinθ)+(1+sinθ)]/(1+sinθ)(1-sinθ) = 2/(1-sin²θ)= 2/cos²θ = 2sec²θ=RHS

Prove each of the following identities: cos²θ+1/(1+cot²θ)=1

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જવાબ : LHS= cos²θ+1/(1+cot²θ) = cos²θ +1/cosec²θ =cos²θ+sin²θ=1= RHS

Prove each of the following identities: tan²θ-1/cos²θ=-1

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જવાબ : LHS =tan²θ – 1/cos²θ = sin²θ/cos²θ – 1/cos²θ = sin²θ-1/cos²θ= -cos²θ/cos²θ = -1= RHS

Prove each of the following identities: cot²θ-1/sin²θ=-1

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જવાબ : LHS=cot²θ -1sinθ =cos²θsin²θ- 1/sin²θ =cos+θ- 1/sin²θ =-sin²θ/sin²θ= -1 =RHS

Prove that, cosec θ (1 + cos θ) (cosec θ − cot θ) = 1

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જવાબ : LHS= cosecθ(1+cosθ)(cosecθ – cotθ ) = (cosecθ + cosecθ × cosθ )( cosecθ – cotθ ) = (cosecθ + cosθ sinθ)(cosecθ−cotθ) = (cosecθ+cotθ)(cosecθ−cotθ) = cosec²θ−cot²θ (∵cosec2θ−cot2θ=1) =1 =RHS

Prove that, (1 + cos θ) (1 − cos θ) (1 + cot²θ) = 1

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જવાબ : LHS=(1+cosθ)(1−cosθ)(1+cot²θ) =(1−cos²θ)cosec²θ =sin²θ×cosec²θ =sin²θ × 1/sin2θ =1 =RHS

Prove that, 1/(1+tan²θ)+1/(1+cot²θ)=1

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જવાબ : LHS=1/(1+tan²θ)+1/(1+cot²θ) =1/sec²θ + 1/cosec²θ = cos²θ+sin²θ =1 =RHS

Prove that, sin²θ+1(1+tan²θ)=1

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જવાબ : LHS= sin²θ + 1/(1+tan²θ) =sin²θ+1/sec²θ (∵sec²θ−tan2θ=1) =sin2θ+cos2θ =1 =RHS

Prove that, (1− cos²θ) sec²θ = tan²θ

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જવાબ : LHS=(1−cos²θ)sec²θ =sin²θ×sec²θ (∵sin²θ+cos²θ=1) =sin²θ× 1/cos²θ = sin²θ/cos²θ =tan²θ =RHS

Prove that, (sec²θ − 1) (cosec²θ − 1) = 1

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જવાબ : LHS= (sec²θ−1)(cosec2θ−1) =tan²θ ×cot²θ (∵sec²θ−tan²θ=1 and cosec²θ−cot²θ=1) =tan²θ × 1/tan+θ =1 = RHS

Prove that, (sec²θ − 1) cot²θ = 1

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જવાબ : LHS= (sec²θ−1)cot²θ = tan²θ×cot²θ (∵sec²θ−tan²θ=1) =1/cot²θ × cot²θ =1 =RHS

Prove that, (1 + cot²θ) sin²θ = 1

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જવાબ : LHS= (1+cot²θ)sin²θ =cosec²θ sin²θ (∵cosec²θ−cot²θ=1) =1/sin²θ × sin²θ =1 Hence, LHS=RHS

Prove that, (1 − cos²θ) cosec²θ = 1

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જવાબ : LHS=(1−cos²θ)cosec²θ =sin²θ cosec²θ (∵cos²θ+sin²θ=1) =1/cosec²θ × cosec²θ =1 Hence, LHS = RHS

Without using trigonometric tables, prove that: (sin 65° + cos 25°)(sin 65° − cos 25°) = 0

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જવાબ : LHS = (sin65°+cos25°)(sin65°−cos25°) = sin²65°−cos²25° = sin²(90°−25°) −cos²25° =cos²25°−cos²25° =0 =RHS

Without using trigonometric tables, prove that: cos²57° − sin²33° = 0

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જવાબ : LHS = cos²57°−sin²33° =cos²(90°−33°) −sin²33° = sin²33°−sin²33° =0 =RHS

Without using trigonometric tables, prove that: sin²48° + sin²42° = 1

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જવાબ : LHS=sin²48° + sin²42° =sin²(90°−42°) +sin²42° =cos²42⁰ + sin²42° =1 =RHS

Without using trigonometric tables, prove that: tan²66° − cot²24° = 0

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જવાબ : LHS = tan²66° − cot²24° =tan²(90°−24°) − cot²24° =cot²24°− cot²24° = 0 =RHS

Without using trigonometric tables, prove that: cos²75° + cos²15° = 1

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જવાબ : LHS = cos²75° + cos²15° =cos²(90°−15°) + cos²15° =sin²15°+cos²15° =1 =RHS

Without using trigonometric tables, prove that: cosec²72° − tan²18° = 1

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જવાબ : LHS = cosec²72° −tan²18° =cosec²(90°−18°) −tan²18° =sec²18° − tan²18° =1 =RHS

Without using trigonometric tables, prove that: cosec 80° − sec 10° = 0

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જવાબ : LHS = cosec80°−sec10° =cosec(90°−10°) −sec10° =sec10°−sec10° = 0 = RHS

Without using trigonometric tables, prove that: tan 71° − cot 19° = 0

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જવાબ : LHS = tan71°−cot19° = tan(90°−19°) − cot19°= cot19°−cot19° =0 = RHS

Without using trigonometric tables, prove that: cos 81° − sin 9° = 0

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જવાબ : LHS= cos81° − sin9° =cos(90°−9°) −sin9° = sin9°−sin9° = 0 =RHS

Without using trigonometric tables,cot 38°/tan 52°

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જવાબ : cot38°/tan52° =cot(90°−52°)/tan52° =tan52°/tan52° =1 [∵tan (90−θ) = cot θ]

Without using trigonometric tables, cosec 42°/sec 48°

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જવાબ : cosec42°/sec48° =cosec(90°−48°)/sec48° =sec48°/sec48° =1 [∵sec (90−θ) = cosec θ]

Without using trigonometric tables, evaluate :cos 35°/sin 55°

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જવાબ : cos35°/sin55° = cos(90°−55°)/sin55° = sin55°/sin55° =1 [∵sin (90−θ) = cosθ]

Without using trigonometric tables, evaluate: tan 27°/cot 63°

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જવાબ : tan27°/cot63° =tan(90°−63°)/cot63°=cot63°cot63° =1 [∵tan (90−θ) = cot θ]

If sinθ = cos(θ-45°), where θ is acute, then find the value of θ.

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જવાબ : We have, sinθ=cos(θ-45°)

⇒cos(90°-θ)=cos(θ-45°)

Comparing both sides, we get

90°- θ = θ - 45°

⇒θ+θ = 90°+45°

⇒2θ=135°

⇒θ=(135/2)°

∴ θ=67.5°

If tanA=5/12, then find the value of (sinA+cosA)secA. [CBSE 2008]

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જવાબ : (sinA+cosA)secA=(sinA+cosA)/cosA=sinA/cosA+cosA/cosA=tanA+1= 5/12+1/1 =(5+12)/12 = 17/12

Write the value of cos1° cos2° ... cos180°.

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જવાબ : cos1° cos2° ... cos180°=cos1° cos2° ... cos90° ... cos180°=cos1° cos2° ... 0 ... cos180°=0

Find the value of sin50°/cos40° + cosec40°/sec50°- 4cos50° cosec40°.

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જવાબ : sin50°/cos40° + cosec40°/sec50° -4cos50° cosec40° = cos(90°-50°)/cos40° + sec(90°-40°)/sec50° - 4sin(90°-50°) cosec40°

=cos40°/cos40° + sec50°/sec50° - 4sin40°/sin40° = 1+1 – 4 = -2

Find the value of sin48° sec42°+cos48° cosec42°.

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જવાબ : sin48° sec42°+cos48° cosec42°= sin48° cosec(90°-42°) + cos48° sec(90°-42°) =sin48° cosec48°+cos48° sec48° =sin48°/sin48° + cos48°/cos48° =1+1 = 2

If sec2A = cosec(A - 42°), where 2A is an acute angle, then find the value of A. [CBSE2008]

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જવાબ : We have,

sec2A = cosec(A−42°)

⇒cosec(90°−2A)=cosec(A−42°)

Comparing both sides, we get

90°−2A = A−42°

⇒2A+A = 90°+42°

⇒3A=132°

⇒A=132°/3

∴ A=44°
Hence, the value of A is 44°.

If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.

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જવાબ : sin3A=cos(A−26°)

⇒cos(90°−3A)=cos(A−26°) [∵sinθ=cos(90°−θ)]

⇒90°−3A=A−26°

⇒116°=4A

⇒A=116°/4 = 290

If tan 2 A = cot (A − 12°), where 2 A is an acute angle, find the value of A.

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જવાબ : tan2A=cot(A−12°)

=>cot(90°−2A)=cot(A−12°) [∵tanθ=cot(90°−θ)]

=>(90°−2A)=(A−12°)

=>102°=3A

=>A=102°/3=34°

If sec 4 A = cosec (A − 15°), where 4 A is an acute angle, find the value of A.

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જવાબ : sec4A=cosec(A−15°)

=> cosec(90°−4A)=cosec(A−15°) [∵secθ=cosec(90°−θ)]

=>90°−4A=A−15°

=>105°=5A

=>A=105°/5 = 21°

tan13°

tan37°

tan45°

tan53°

tan77°

= -1

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જવાબ : tan13° tan37° tan45° tan53° tan77°

tan13°

tan37°

cot(90° - 53°)

cot(90° -77°

) 1

tan13°

tan37°

cot37°

cot13°

- 5/3

=2/3 – 5/3

=−1
Hence Proved

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find AB.

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/20Q(14).png

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જવાબ : AB/AC = cos 30°

⇒AB/20 = √3/2

⇒AB = (20×√3/2 ) = 10√3cm

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find BC, (ii) AB.

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જવાબ : From the given right-angled triangle, we have:
BC/AC = sin 30°

⇒BC/20 = 1/2

⇒BC = 20/2 = 10 cm

In the adjoining figure, ∆ABC is a right-angled at B and ∠A = 45°. If AC = 32–√32 cm,
find AB.

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/22(182).png

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જવાબ : From right-angled ∆ABC, we have:

AB/AC=cos 45°

⇒AB/3√2 =1/√2

⇒ AB=3 cm

In the adjoining figure, ∆ABC is a right-angled at B and ∠A = 45°. If AC = 32–√32 cm,
find BC, (ii) AB.

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જવાબ : From right-angled ∆ABC, we have:

BC/AC = sin 45°

⇒BC/3√2=1/√2

⇒ BC=3 cm

If sin(A+B)=sinA cosB+cosA sinB and cos(A−B)=cosA cosB+sinA sinB , find the values of cos15°

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જવાબ : cos(A−B)=cosA cosB+sinA sinB

⇒cos(45°−30°)=cos45°cos30°+sin45°sin30°

⇒cos(15°) =

⇒cos15°=

∴ cos15°=

If sin(A+B)=sinA cosB+cosA sinB and cos(A−B)=cosA cosB+sinA sinB , find the values of sin75°

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જવાબ : Let A=45° and B=30°Let A=45° and B=30°
sin(A+B)=sinA cosB+cosA sinB

⇒sin(45°+30°)=sin45° cos30°+cos45° sin30°

⇒sin(75°)=

⇒sin75°=

∴ sin75°=

If 3x=cosecθ and 3x=cotθ , find the value of 3(x²−1/x²). [CBSE 2010]

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જવાબ : 3(x²−1/x²) = = = = = 1/3

If tan (A − B) = 1/√3 and tan (A + B) = √3, 0° < (A + B) < 90° and A > B, then find A and B.

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જવાબ : Here, tan (A − B) =  1/√3
⇒ tan (A − B) = tan 30^o   [∵ tan 30^o = 1/√3 ]
⇒ A − B = 30^o                     ...(A)

Also, tan (A + B) = √3
⇒ tan (A + B) = tan 60^o      [∵ tan 60^o = √3]
⇒ A + B = 60^o...(B)

Solving (A) and (B), we get:
A = 45^o and B = 15^o

If sin (A − B) = ½ and cos (A + B) = ½, 0° < (A + B) < 90° and A > B, then find A and B.

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જવાબ : Here, sin (A − B) = 1212
⇒ sin (A − B) = sin 30^o      [∵ sin 30^o = 1212]
⇒ A − B = 30^o                                  ...(A)

Also, cos (A + B) = 1212
⇒ cos (A + B) = cos 60^o        [∵ cos 60^o = 1212]
⇒ A + B = 60^o                                 ...(B)

Solving (A) and (B), we get:
A = 45^o and B = 15^o

If sin (A + B) = 1 and cos (A − B) = 1, 0° ≤ (A + B) ≤ 90° and A > B, then find A and B.

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જવાબ : Here, sin (A + B) = 1
⇒ sin (A+ B) = sin 90^o   [∵ sin 90^o = 1]
⇒ A + B = 90^o                       ...(A)

Also, cos (A − B) = 1
⇒ cos (A − B) = cos 0^o      [∵ cos 0^o = 1]
⇒ A − B = 0^o                      ...(B)

Solving (A) and (B), we get:
A = 45^o and B = 45^o

There are No Content Availble For this Chapter

1	Sin A	A	AB/AC
2	Cos A	B	BC/AC
3	Tan A	C	BC/AB

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જવાબ :

1-B, 2-A, 3-C

1	Cot A	A	AC/AB
2	Cosec A	B	AB/BC
3	Sec A	C	AC/BC

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જવાબ :

1-B, 2-C, 3-A

1	Sin 0°	A	32
2	Sin 30°	B	0
3	Sin 45°	C	½
4	Sin 60°	D	1/√2

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જવાબ :

1-B, 2-C, 3-D, 4-A

1	Sin θ	A	Perpendicular ÷ Base
2	Cos θ	B	Perpendicular ÷ Hypotenuse
3	Tan θ	C	Base ÷ Hypotenuse

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જવાબ :

1-B, 2-C, 3-A

1	Cos 0°	A	½
2	Cos 30°	B	1/√2
3	Cos 45°	C	32.
4	Cos 60°	D	1

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જવાબ :

1-D, 2-C, 3-B, 4-A

1	Cot 60°	A	2
2	Sec 60°	B	2/
3	Cosec 60°	C	1/.

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જવાબ :

1-C, 2-A, 3-B

1	Cot 30°	A	2/
2	Cosec 30°	B	3
3	Sec 30°	C	2

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જવાબ :

1-B, 2-C, 3-A

1	Sin 90°	A	NA.
2	Cos 90°	B	1
3	Tan 90°	C	0

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જવાબ :

1-B, 2-C, 3-A

1	Cot θ	A	Hypotenuse ÷ Base
2	Cosec θ	B	Base ÷ Perpendicular
3	Sec θ	C	Hypotenuse ÷ Perpendicular

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જવાબ :

1-B, 2-C, 3-A

1	Tan 0°	A	1
2	Tan 30°	B	.
3	Tan 45°	C	1/
4	Tan 60°	D	0

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જવાબ :

1-D, 2-C, 3-A, 4-B

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Introduction to Trigonometry

Math
Chapter 08 : Introduction to Trigonometry

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Introduction to Trigonometry

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