# CBSE Solutions for Class 10 English

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

cos 60° cos 30° − sin 60° sin 30° = ?

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

Evaluate:

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

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° = ?

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

cos 45° cos 30° + sin 45° sin 30° = ?

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

Evaluate:

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

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° = ?

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

જવાબ : cot2 30o − 2 cos2 30o – 3/4sec2 45o + 1/4cosec230o
= (√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°) = ?

જવાબ : (sin2 30o + 4 cot2 45o − sec2 60o )(cosec2 45sec2 30o)
=[(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°-sin2 = ?

જવાબ : 4cot230° +1sin230° -2 cos245o -sin20o =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

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°

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°

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

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

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

Verify:

2 sin 30° cos 30° = sin 60°

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

Verify:

2 sin 45° cos 45° = sin 90°

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

2 sin 45o cos 45o = sin 90o

If A = 30°, verify that: tan2A =

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

If A = 30°, verify that: cos2A =

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

If A = 30°, verify that:  sin2A=

જવાબ : A = 30o
2A = 2 × 30o = 60o
sin 2A = sin 60o = √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

જવાબ : A = 45o
⇒ 2A = 2 × 45o = 90o
cos 2A = cos 90o = 0
2 cos2 A − 1 = 2 cos2 45o − 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 cos A

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

sin 2A = 2 sin A cos A

tan2θ/(1+tan2θ) + cot2θ/(1+cot2θ) = 1

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

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

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

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

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

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

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

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

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

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

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

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

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

Prove that, sin2θ+1(1+tan2θ)=1

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

Prove that, (1− cos2θ) sec2θ = tan2θ

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

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

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

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

Prove that, (1 + cot2θ) sin2θ = 1

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

Prove that, (1 − cos2θ) cosec2θ = 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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]

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°.

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

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

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°.

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]

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 is 44°.

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

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.

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.

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

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

= tan13° tan37° cot(90° - 53°) cot(90° -77°) 1

= tan13° tan37° cot37° cot13° 1

- 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.

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.

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.

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.

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°

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°

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]

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.

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

Also, tan (A + B) = √3

​ tan (A + B) =  tan 60o        [ tan 60o = √3]
A + B = 60o                           ...(B)

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

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

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

Also, cos (A + B) = 1212

​ cos (A + B) =  cos 60o              [​ cos 60o = 1212]
A + B = 60o                                 ...(B)

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

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

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

Also, cos (A − B) = 1

​ cos (A − B) = cos 0o                    [​ cos 0o = 1]
A − B = 0o                        ...(B)

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

### 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

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

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

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

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

જવાબ :

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

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

જવાબ :

1-C, 2-A, 3-B

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

જવાબ :

1-B, 2-C, 3-A

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

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

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

જવાબ :

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