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\(\frac{cos^2θ}{sin θ}-cosec θ+sin θ \) = \(\frac{cos^2θ} {sin θ}-\frac{1}{sin θ}+sinθ\)

=  \(\frac{cos^2θ-1+sin^2θ} {sin θ}\)  

= \(\frac{(cos^2θ+sin^2θ)-1}{sinθ}\) 

= \(\frac{1-1}{sin θ}\)  =0

Hence Proved.

\(\frac{cosθ}{cosecθ+1}+\frac{cosθ}{cosecθ-1}\)

= \(\frac{cosθ(cosecθ-1)+cosθ(cosecθ+1)}{(cosecθ+1)(cosecθ-1)}\) 

= \(\frac{2cotθ}{cot^2θ}\) 

= 2 tanθ

Hence Proved.

 \(\sqrt{\frac{1-cosθ}{1+cosθ}}\) = \(\sqrt{\frac{1-cosθ}{1+cosθ}\times\frac{1-cosθ}{1-cosθ}}\)

= \(\sqrt{\frac{(1-cosθ)^2}{1-cos^2θ}}\)

=  \(\sqrt{\frac{(1-cosθ)^2}{sin^2θ}}\)

= \(\frac{1-cosθ}{sinθ}\)

= \(\frac{1}{sinθ}-\frac{cosθ}{sinθ}\)

= cosecθ-cotθ

Hence Proved.

Taking RHS 

tan⁶θ + 3tan²θsec²θ + 1 

= (sec²θ - 1)³ + 3(sec²θ - 1)sec²θ + 1 

[As, tan²θ = sec²θ - 1] 

= (sec⁶θ - 1 - 3sec⁴θ + 3sec²θ) + (3sec⁴θ - 3sec²θ) + 1 

[(a + b)³ = a³ - b³ - 3a²b + 3ab²] 

= sec⁶θ= LHS Hence Proved.

To Prove : (1-cos²A)cosec²A =1

Proof:

1 - cos²A = = sin² A

Therefore, L.H.S = sin²A .cosec² A

Now, cosec²A = \(\frac{1}{sin^2A}\)

Therefore, L.H.S = sin²A .   \(\frac{1}{sin^2A}\)= 1

R.H.S = 1

L.H.S = R.H.S

Hence Proved.

To prove:  sin²A cos²B - cos²Asin²B = sin²A - sin²B

Proof: Take LHS, Use the identity sin²θ+cos²θ=1

sin²A cos²B - cos²A sin²B

= sin²A(1 – sin²B) – (1 – sin²A)sin²B 

sin²B – sin²B + sin²A sin²B

= sin²A – sin²B

= RHS Hence Proved

cot²A cosec²B - cot²B cosec²A 

= cot²A(1+ cot²B) - cot²B(1+ cot²A)

= cot²A + cot²A cot²B - cot²B - cot²A cot²B

= cot²A - cot²B

Hence Proved.

 cos²A + \(\frac{1}{1+cot^2A}=cos^2A+\frac{1}{cosec^2A}\)

= cos²A + sin²A

= 1

Hence Proved.

The correct option is (c) 90.

(cosec θ + cot θ) = m …(1) and

(cosec θ − cot θ) = n …(2)

Multiply (1) and (2)

(cosec² θ – cot² θ) = mn

(Because cosec² θ – cot² θ = 1)

1 = mn

Or mn = 1

Hence Proved

√3tanθ = 3sinθ

√3\(\frac{sinθ}{cosθ}\) = √3 x √3sinθ

\(\frac{1}{cosθ}\) = √3

cosθ = \(\frac{1}{\sqrt{3}}\) 

Now, sin²θ - cos²θ = 1 - cos²θ - cos²θ

= 1 - 2 cos²θ

= 1 - 2 x \(\Big(\frac{1}{\sqrt{3}}\Big)^2\) 

= 1 - 2 x \(\frac{1}{3}=\frac{1}{3}\)

\(\frac{(1+tan^2θ)cotθ}{cosec^2θ}\)   =   \(\frac{sec^2\times θcotθ}{cosec^2θ}\)

=  \(\frac{sin^2θ}{cos^2θ}\times cotθ\)

 = tan²θ x cotθ

 = tan²θ x \(\frac{1}{tanθ}\)

 = tanθ 

Hence Proved.

Given: sec²θ (1 + sin θ) (1 − sin θ) = k 

To find: k 

Consider sec²θ (1 + sin θ) (1 − sin θ) 

∵ (a – b) (a + b) = a² – b² 

∴ sec²θ (1 + sin θ) (1 − sin θ) = sec²θ(1 – sin²θ) 

Now, as sin²θ + cos²θ = 1 

⇒ cos²θ = 1 – sin²θ 

⇒ sec²θ(1 + sin θ)(1 − sin θ) = sec²θ(1 – sin²θ) 

= sec²θ cos²θ 

Now, 

∵  secθ = \(\frac{1}{cosθ}\)

⇒ sec²θ = \(\frac{1}{cos^2θ}\)

⇒ sec²θ(1 + sin θ)(1 − sin θ) = sec²θ(1 – sin²θ) 

= sec²θ cos²θ 

= \(\frac{1}{cos^2θ }\)cos²θ = 1

⇒ k = 1

Given: sin²θ cos²θ (1 + tan²θ)(1 + cot²θ) = λ 

To find: λ 

We know that 1 + tan²θ = sec²θ 

And 1 + cot²θ = cosec²θ 

⇒ sin²θ cos²θ (1 + tan²θ) (1 + cot²θ) 

= sin²θ cos²θ sec²θ cosec²θ

Now,

∵ cosecθ = \(\frac{1}{sinθ}\)

⇒ cosec²θ = \(\frac{1}{sin^2θ}\)

And ∵ secθ = \(\frac{1}{cosθ}\)

⇒ sec²θ = \(\frac{1}{cos^2θ}\)

⇒ sin²θ cos²θ (1 + tan²θ) (1 + cot²θ)

= sin²θ cos²θ sec²θ cosec²θ

= sin²θ cos²θ \(\frac{1}{cos^2θ}\)\(\frac{1}{sin^2θ}\)= 1

⇒ λ = 1

Given: cosθ = \(\frac{3}{4}\)

To find: 9 tan²θ + 9 

∵ secθ = \(\frac{1}{cosθ}\) = \(\frac{4}{3}\)

∴ sec²θ = \(\Big(\frac{4}{3}\Big)^2\)= \(\frac{16}{9}\)

Also, we know that 1 + tan²θ = sec²θ

⇒ tan²θ = sec²θ - 1

=\(\frac{16}{9}\) - 1 = \(\frac{16-9}{9}\) = \(\frac{7}{9}\)

⇒ 9tan²θ + 9 = 9\(\Big(\frac{7}{9}\Big)\) + 9 = 7 + 9 = 16

Given: sinθ = \(\frac{1}{3}\)

To find: The value of 2cot²θ + 2. 

Solution: \(\)​sinθ = \(\frac{1}{3}\)

∵ cosecθ = \(\frac{1}{sinθ}\) = \(\frac{1}{1/3}\) = 3

⇒ cosec²θ = 3² = 9

Also, 1 + cot²θ = cosec²θ 

⇒ cot²θ = cosec²θ – 1 = 9 – 1 = 8 

⇒ 2 cot²θ + 2 = 2 (8) + 2 = 16 + 2 = 18

Hence, 

the value of 2cot²θ + 2 is 18.

To find: 6tan²θ - \(\frac{6}{cos^2θ}\) 

∵ secθ = \(\frac{1}{cosθ}\) 

⇒ sec²θ = \(\frac{1}{cos^2θ}\)

⇒6tan²θ - \(\frac{6}{cos^2θ}\) = 6tan²θ -6sec²θ = 6(tan²θ - sec²θ)

Now, as 1 + tan²θ = sec²θ 

⇒ tan²θ – sec²θ = – 1

⇒ 6tan²θ - \(\frac{6}{cos^2θ}\) = 6(tan²θ - sec²θ) = 6(-1) = - 6

To find: 9 cot²θ – 9 cosec²θ

Consider 9 cot²θ – 9 cosec²θ = 9 (cot²θ – cosec²θ) 

Now 

1 + cot² θ = cosec²θ 

⇒ cot²θ – cosec²  = – 1 

⇒ 9 cot²θ – 9 cosec²θ = 9 (cot²θ – cosec²θ) 

= 9 ( – 1) = – 9

To find: (1 + tan²θ) (1 − sin θ) (1 + sin θ) 

∵ (a – b) (a + b) = a² – b² 

∴ (1 + tan²θ) (1 − sin θ) (1 + sin θ) 

= (1 + tan²θ) (1 – sin²θ) 

Now, as sin²θ + cos²θ = 1 

⇒ 1 – sin²θ = cos²θ………(i)

Also, we know that 1 + tan²θ = sec²θ ……(ii) 

Using (i) and (ii), we have 

(1 + tan²θ) (1 − sinθ) (1 + sinθ) 

= (1 + tan²θ) (1 – sin²θ) 

= sec²θ cos²θ

∵secθ = \(\frac{1}{cosθ}\) 

sec²θ = \(\frac{1}{cos^2θ}\) 

⇒ (1 + tan²θ) (1 − sinθ) (1 + sinθ)

= sec²θ cos²θ

= \(\frac{1}{cos^2θ}{cos^2θ}= 1\)

L.H.S. = sin θ (1 + tan θ) + cos θ (1+ cot θ)

= sin θ (1 + sin θ/cos θ) + cos θ (1+ cos θ/sinθ)

= sin θ{(cosθ +sinθ )/cos θ} + cos θ{(sinθ+cosθ )/sinθ}

=(cos θ + sin θ) (sinθ/cos θ + cos θ/sinθ)

= (cos θ + sin θ)/cos θ sin θ

= cosec θ + sec θ

= R.H.S.

To find:  \(\frac{tan^2θ - sec^2θ}{cot^2θ - cosec^2θ}\)

We know that 1 + tan²θ = sec²θ 

And 1 + cot²θ = cosec²θ 

⇒ tan²θ – sec²θ = – 1 

And cot²θ – cosec²θ = – 1

⇒ \(\frac{tan^2θ - sec^2θ}{cot^2θ - cosec^2θ}\) = \(\frac{-1}{-1}\) = 1

4tan²θ – 4/cos²θ 

= 4 x sin²θ/cos²θ – 4/cos²θ

= (4(sin²θ – 1))/cos²θ

= 4 (-cos²θ) / cos²θ

= -4

(tan²θ – sec²θ) / (cot²θ – cose²θ) 

= -1/-1

= 1

(Using 1 + cot²θ = cose²θ and 1 + tan²θ = sec²θ)

Given: a cosθ + b sinθ = 4

Squaring both sides, we get 

(a cosθ + b sinθ)² = 4² 

⇒ a² cos²θ + b² sin²θ + 2ab sin θ cos θ = 16 …(i) 

and a sinθ – b cosθ = 3 

Squaring both sides, we get 

(a sinθ – b cosθ)² = 3² 

⇒ a² sin²θ + b² cos²θ – 2ab sinθ cosθ = 9 …(ii) 

To find: a² + b² 

Adding (i) and (ii), we get 

a² cos²θ + b² sin²θ + 2ab sinθ cosθ + a² sin²θ + b²cos²θ – 2ab sinθ cosθ = 16 + 9

 ⇒ a² (sin²θ + cos²θ) + b² (sin² θ + cos²θ) = 25 

⇒ a² + b² = 25 [∵ sin²θ + cos²θ = 1]

Given: a cot θ + b cosec θ = p 

Squaring both sides, we get 

(a cot θ + b cosec θ)² = p²

⇒ a² cot²θ + b² cosec²θ + 2ab cotθ cosecθ = p² ……(i) 

and b cotθ + a cosecθ = q 

Squaring both sides, we get 

(b cot θ + a cosec θ)² = q² 

⇒ b² cot²θ + a² cosec²θ + 2ab cotθ cosecθ = q² ……(ii) 

To find: p² – q² 

Subtracting (ii) from (i), we get 

a² cot²θ + b² cosec²θ + 2ab cotθ cosecθ – b² cot²θ – a² cosec²θ – 2ab cotθ cosecθ = p² – q² 

⇒ P² – q² = a² (cot²θ – cosec²θ) + b² (cosec²θ – cot²θ) 

= a² ( – 1) + b² (1) [∵1 = cosec²θ – cot²θ] 

= b² – a²

To find: sin² 29° + sin²61° 

Consider sin² 29° + sin²61° 

∵ 29 = 90 – 61 

∴ sin² 29° + sin²61° = sin² (90° – 61°) + sin² 61° 

Now, as sin (90° – θ) = cos θ 

⇒ sin² 29° + sin²61° = sin² (90° – 61°) + sin2 61° 

= cos² 61° + sin² 61° 

= 1 [sin²θ + cos²θ = 1]

Consider, 

(1+cot²A)sin²A 

As we know 1+cot²A = cosec²A 

Putting the values we get, 

(cosec²A)sin²A 

As we know,cosec A = 1/sinA 

So,

\(\frac{1}{sin^2A}\times\,sin^2A\) = 1

hence proved

Given : sin² A cot²A + cos²A tan²A =1

To prove : Above equality holds. Proof: Consider LHS, we know,

cot θ =  \(\frac{cosθ}{sinθ}\)   and  tanθ = \(\frac{sinθ}{cosθ}\)

using these  

sin²A cot²A + cos²A tan²A

= sin²A x \(\frac{cos^2A}{sin^2A}\)  + cos²x \(\frac{sin^2A}{cos^2A}\)  

= cos²A sin²A

= 1

Which is equal to RHS. 

Hence Proved

Given: sec θ + tan θ = x ...…(i) 

To find: tan θ 

We know that 1 + tan²θ = sec²θ 

⇒ sec²θ – tan²θ = 1 

∵ a² – b² = (a – b) (a + b) 

∴ sec²θ – tan²θ = (secθ – tanθ) (secθ + tanθ) = 1 

⇒ From (i), we have 

⇒ (secθ – tanθ) x = 1 

⇒ secθ – tanθ = \(\frac{1}{X}\)………(ii) 

Subtracting (ii) from (i), we get

tanθ + tanθ = \(X-\frac{1}{X}\) 

2tanθ = \(\frac{X^2-1}{X}\) 

tanθ = \(\frac{X^2-1}{2X}\)

Given: secθ + tanθ = x …(i) 

To find: secθ 

We know that 1 + tan²θ = sec²θ 

⇒ sec²θ – tan²θ = 1 

∵ a² – b² = (a – b) (a + b) 

∴ sec²θ – tan²θ = (secθ – tanθ) (secθ + tanθ) = 1

 ⇒ From (i), we have 

⇒ (secθ – tanθ) x = 1 

⇒ secθ – tanθ = \(\frac{1}{x}\) ...…(ii) 

Adding (i) and (ii), we get

secθ + secθ = \(X+\frac{1}{X}\) 

⇒ secθ =  \(\frac{X^2+1}{2X}\) 

⇒ secθ = \(\frac{X^2+1}{2X}\)

Given: cosecθ = 2x 

⇒ x = \(\frac{ cosec θ}{2}\) 

⇒ x² = \(\frac{ cosec^2θ}{4}\)  .........(1)

And cotθ = \(\frac{2}{x}\) 

⇒ x =  \(\frac{2}{cotθ}\) 

⇒ x² =  \(\frac{4}{cot^2θ}\) 

⇒ \(\frac{1}{x^2}\) = \(\frac{cot^2θ}{4}\) .....(ii)

To find: \(2\Big(x^2-\frac{1}{x^2}\Big)\) 

Consider \(2\Big(x^2-\frac{1}{x^2}\Big)\) = \(2\Big(\frac{cosec^2 θ}{4}-\frac{1}{x^2}\Big)\) [Using (i)]

= \(2\Big(\frac{cosec^2 θ}{4}-\frac{cot^2 θ}{4}\Big)\) [Using (ii)]

= \(2\Big(\frac{cosec^2 θ-cot^2 θ}{4}\Big)\) = \(\frac{1}{2}\) (cosec²θ – cot²θ)

Now, as 1 + cot²θ = cosec²θ 

⇒ 1 = cosec² θ – cot² θ

⇒ \(2\Big(x^2-\frac{1}{x^2}\Big)\) = \(\frac{1}{2}\) (cosec²θ – cot²θ) = \(\frac{1}{2}\)

cosecθ + cotθ = m

cosecθ + cotθ = n

mn = (cosecθ + cotθ)(cosecθ - cotθ)

mn = (cosec²θ + cot²θ)

mn = 1

Hence Proved.

sinθ + cos θ = √2cos(90° - θ)

cos θ = √2sinθ - sinθ

cos θ = (√2-1)sinθ

\(\frac{cosθ}{sinθ}\) = √2-1

\(\frac{cosecA}{cosecA-1}+\frac{cosecA}{cosecA+1}\) 

= \(\frac{cosecA(cosecA+1)+cosecA(cosecA-1)}{cosec^2A-1}\) 

= \(\frac{cosec^2A+cosecA+cosec^2A-cosecA}{cosec^2A-1}\) 

= \(\frac{2cosec^2A}{cot^2A}\) = \(\frac{2}{sin^2A}\times \frac{sin^2A}{cos^2A}\) = 2sec²A

Hence Proved.

\(\frac{1+sec θ}{sec θ}\)  = \(1+\frac{1}{\frac{cos θ}{\frac{1}{cos θ}}}\)

= \(\frac{1+cos θ}{1}\)  

= \( \frac{(1+cos θ)(1-cos θ)}{(1-cos θ)}\)  

= \( \frac{1-cos^2 θ}{(1-cos θ)}\) 

= \(\frac{sin^2 θ}{1-cos θ}\) 

Hence Proved.