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sec A(1 - sin A)(sec A + tan A)

 \(\Big(sec A - \frac{1}{cos A}\times sinA\Big)(sec A + tan A)\)                       

(sec A - tan A)(sec A + tan A)

(sec² A + tan² A)

(tan² A + 1 - tan² A)

= 1

Hence Proved.

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

= \(\frac{1-cos^2θ}{sinθ(1+cosθ)}\)

= \(\frac{sin^2θ}{sinθ(1+cosθ)}\)

= \(\frac{sinθ}{1+cosθ}\)​

​​​​​​Hence Proved.

sinθ  = \(\frac{3}{5}\) ,tanθ = \(\frac{3}{4}\) ,secθ = \(\frac{5}{4}\) ,

cosθ = \(\frac{4}{5}\) ⇒ sinθ = \(\frac{3}{5}\) 

secθ = \(\frac{1}{cosθ}\) = \(\frac{1}{4/5}\) = \(\frac{5}{4}\) 

tanθ = \(\frac{sinθ}{cosθ}\) = \(\frac{3/5}{4/5}\) = \(\frac{3}{4}\) 

Hence Proved.

(i) sin 81° + sin 71°

= sin(90° – 9°) + sin(90° – 19°)

= cos 9° + cos 19°

(ii) tan 68° + sec 68°

= tan(90° – 22°) + sec(90° – 22°)

= cot 22° + cosec 22°

We have, 

cos θ = \(\frac{4}{5}\) 

And we know that, 

sin θ = √(1 – cos² θ)

⇒ sin θ = √(1 – (\(\frac{4}{5}\))²) 

= √(1 – (\(\frac{16}{25}\))) 

= √[\(\frac{(25 \,–\, 16)}{25}\)] 

= √(\(\frac{9}{25}\)) 

= \(\frac{3}{5}\)

∴ sin θ =\(\frac{3}{5}\)

Since, 

cosec θ = \(\frac{1}{sin\, θ }\)

= \(\frac{1}{(3/5)}\)

⇒ cosec θ = \(\frac{5}{3}\) 

And, sec θ = \(\frac{1}{cos θ }\)

= \(\frac{1}{(4/5)}\)

⇒ cosec θ = \(\frac{5}{4}\) 

Now, 

tan θ = \(\frac{sin θ}{cos θ }\)

= \(\frac{(3/5)}{(4/5)}\)

⇒ tan θ = \(\frac{3}{4}\) 

And, cot θ = \(\frac{1}{tan θ}\) 

= \(\frac{1}{(3/4)}\)

⇒ cot θ = \(\frac{4}{3}\)

To find: cos⁴A – sin⁴A 

Consider cos⁴A – sin⁴A = (cos²A)² – (sin²A)² 

∵ a² – b² = (a – b) (a + b)

∴ cos⁴A – sin⁴A = (cos²A)² – (sin2 A)² 

= (cos²A – sin²A) (cos²A + sin²A) 

= (cos²A – sin²A) [∵ cos²A + sin²A = 1] 

= cos²A –(1 – cos²A) [∵ sin²A = 1 – cos²A] 

= cos²A – 1 + cos²A = 2 cos²A – 1

sinθ + cosθ = x

Squaring on both sides

(sinθ + cosθ)² = x²

⇒ sin²θ + cos²θ + 2sinθcosθ = x²

∴ sinθcosθ = \(\frac{x^2-1}{2}\) ...(1)

We know sin²θ + cos²θ = 1

combining both since

(sin²θ + cos²θ)³ = (1)³

sin⁶θ + cos⁶θ + 3sinθCos²θ (sin²θ + cos²θ) = 1

⇒ sin⁶θ + cos⁶θ = 1- 3sin²θCos²θ

= 1- \(\frac{3(x^2-1)^2}{4}\) from -(1)

sin⁶θ + cos⁶θ = \(\frac{4-3(x^2-1)^2}{4}\) 

Hence Proved.

We know that, 

cot² A = \(\frac{cos^2 A}{sin^2 A}\) and tan² A = \(\frac{sin^2 A}{cos^2 A}\)

Substituting the above in L.H.S, we get 

L.H.S = sin² A cot² A + cos² A tan² A 

= {sin² A (\(\frac{cos^2 A}{sin^2 A}\))} + {cos² A (\(\frac{sin^2 A}{cos^2 A}\))} 

= cos² A + sin² A = 1 [∵ sin² θ + cos² θ = 1] 

= R.H.S 

Hence Proved

sinθ cos(90°-θ)+cosθ sin(90°-θ) 

= sinθ x sinθ + cosθ x cosθ

= sin²θ + cos²θ

= 1

Consider ΔABC to be a right angled triangle at B.

angle C = 90 degree

Given: tan A = 1 …(1)

tan A = 1 = BC/AB

AB = BC

Again, tan A = sin A/cos A

sin A = cos A …using (1)

By Pythagoras theorem:

AC² = BC² + AB²

AC² = 2BC²

(AC/BC)² = 2

Or AC/BC = √2

cosec A = √2

or sin A = 1/√2

and cos A = 1/√2

Now,

2 sin A cos A = 2(1/√2)( 1/√2)

= 2(1/2)

= 1

= RHS

(sec A - tan A)² = \(\Big(\frac{1}{cosA}-\frac{sinA}{cosA}\Big)^2\)

 = \(\frac{(1-sinA)^2}{cos^2A}\)  

=  \(\frac{(1-sinA)^2}{1-sin^2A}\)  

= \(\frac{(1-sinA)^2}{(1-sinA)(1+sinA)}\)

=  \(\frac{1-sinA}{1+sinA}\)  

Hence Proved.

sec A = 1/cos A

tan A = sin A/ cos A

Thus,

L.H.S=(1/cosA - sin A/cos A)²

=(1-sin A)²/cos² A

=(1-sin A)(1- sin A)/(1-sin² A)                      By a²-b²=(a+b)(a-b) and sin² A + cos² A=1

=(1-sin A)/(1+sin A)=R.H.S(corrected)

Hence, proved

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

  =   \(\sqrt{\frac{(1+sinA)^2}{1-sin^2A}}\)  

= \(\sqrt{\frac{(1+sinA)^2}{cos^2A}}\)  

=  \(\frac{1+sinA}{cosA}\)  

=  \(\frac{1}{cosA}+\frac{sinA}{cosA}\)

= secA + tanA

Hence Proved.

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

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

= \(\frac{[(sinθ+cosθ)-1][(sinθ+cosθ)+1]}{sinθ.cosθ}\) 

= \(\frac{(sinθ+cosθ)^2-(1)^2}{sinθ.cosθ}\) 

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

= \(\frac{1+2sinθcosθ-1}{sinθ.cosθ}\) 

= \(\frac{2sinθcosθ}{sinθ.cosθ}\) = 2 = R.H.S

Hence Proved.

We have, 

L.H.S = tan θ + \(\frac{1}{tan\, θ}\) 

= \(\frac{(tan^2 θ\, +\, 1)}{tan θ}\)

= \(\frac{sec^2 \,θ }{tan \,θ}\) [∵ sec² θ − tan² θ = 1] 

= (\(\frac{1}{cos^2\, θ}\)) x \(\frac{1}{sinθ \over cos θ}​\) [∵ tan θ = \(\frac{sin \,θ }{cos \,θ}\)] 

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

= \(\frac{1}{cos\, θ}\) x \(\frac{1}{sin\, θ}\)

= sec θ x cosec θ 

= sec θ cosec θ 

= R.H.S 

Hence Proved

Using identities, 

(sec² θ − tan² θ) = 1 and (cosec² θ − cot² θ) = 1 

We have, 

L.H.S = (sec² θ – 1)(cosec²θ – 1) 

= tan²θ × cot²θ 

= (tan θ × cot θ)² 

= (tan θ × 1/tan θ)² 

= 1² 

= 1 

= R.H.S 

Hence Proved

(2 sin θ + 3 cos θ) = 2 …(1)

(2 sin θ + 3 cos θ)² + (3 sin θ – 2 cos θ)²

= 4sin² θ + 9 cos² θ + 12sin θ cos θ + 9 sin² θ + 4 cos² θ – 12 sin θ cos θ

= 13sin² θ + 13 cos² θ

= 13(sin² θ + cos² θ)

= 13

(Because (sin² θ + cos² θ) = 1)

=> (2 sin θ + 3 cos θ)² + (3 sin θ – 2 cos θ)² = 13

Using equation (1)

=> (2)² + (3 sin θ – 2 cos θ)² = 13

=> (3 sin θ – 2 cos θ)² = 9

or (3 sin θ – 2 cos θ) = ± 3

Hence Proved.

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

= \(\frac{cosθ(1-sinθ)}{1-sin^2θ}\)

= \(\frac{cosθ(1-sinθ)}{cos^2θ}\)

= \(\frac{1-sinθ}{cosθ}\)

Hence Proved.

\(\frac{1}{secA-1}+\frac{1}{secA+1}\) = \(\frac{secA+1+secA-1}{sec^2A-1}\)  

=   \(\frac{2secA}{tan^2A}\) 

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

= 2cosec A cot A

Hence Proved.

We have

Cot A = 4/3

cot ( 90-B) = 4/3 (A+B = 90)

tan B = 4/3

We have

cos B = 3/5

cos (90- A) = 3/5

sin A = 3/5

sinθ cos(90-θ) + cosθ sin(90-θ)

= sinθsinθ + cosθcosθ

= sin²θ + cos²θ

= 1

To find: (1 – cos²θ) cosec²θ

∵ cosecθ = \(\frac{1}{sinθ}\)

∴ cosec²θ = \(\frac{1}{sin^2θ}\) 

⇒ (1 – cos²θ) cosec²θ = (1 – cos²θ)\(\frac{1}{sin^2θ}\) ....(1)

∵ sin² θ + cos² θ = 1 

∴ sin² θ = 1 – cos² θ 

⇒ from (i), we have

⇒ (1 – cos²θ) cosec²θ = sin² θ \(\frac{1}{sin^2θ}\) = 1

Given: x = a sin θ and y = b cos θ 

⇒ x² = a² sin²θ and y² = b² cos²θ ………(i) 

To find: b²x² + a²y² 

Consider b²x² + a²y² = b² a² sin²θ + a² b² cos²θ 

= a²b² (sin²θ + cos²θ) 

= a²b² (1) [∵ sin²θ + cos²θ = 1] 

= a²b²

(1 – sin²θ) sec² θ 

= (cos² θ) × 1/cos² θ

= 1

To find: (1 + cot²θ) sin²θ 

∵ 1 + cot² θ = cosec² θ 

∴ (1 + cot²θ) sin²θ = cosec²θ sin²θ

Also, cosecθ = \(\frac{1}{sinθ}\) 

⇒ cosec²θ = \(\frac{1}{sin^2θ}\)

(1 + cot²θ) sin²θ = cosec²θ sin²θ = \(\frac{1}{sin^2θ}sin^2θ\) = 1

Given: x = asin θ and y = bcos θ 

⇒ x² = a² sin² θ and y² = b² cos²θ ………(i) 

To find: b²x² + a²y² 

Consider b²x² + a²y² = b²a² sin²θ +a²b²cos²θ 

= a² b² (sin²θ + cos²θ) 

= a² b² (1) [∵ sin² θ + cos²θ = 1] 

= a² b²

To find: sin A cos (90° − A) + cos A sin (90° − A) 

∵ cos (90° – A) = sin A and sin (90° – A) = cos A ……………(i) 

∴ sin A cos (90° − A) + cos A sin (90° − A) 

= sin A sin A + cos A cos A [Using (i)] 

= sin²A + cos²A 

Now, 

∵ sin² θ + cos² θ = 1 

∴ sin A cos (90° − A) + cos A sin (90° − A) 

= sin²A + cos²A = 1

(1-cos²θ)cosec²θ

= sin²θ x 1/sin²θ

= 1

(1 – cos²θ) cosec²θ = 1

L.H.S. = (1 – cos²θ) cosec²θ

= (sin²θ) × cosec²θ

(Using identity sin²θ + cos² θ = 1)

= 1/ cosec²θ × cosec²θ

= 1

= R.H.S.

Hence Proved.

To find: cot²θ - \(\frac{1}{sin^2θ}\) 

∵ cosecθ  = \(\frac{1}{sinθ}\)

cosec²θ  = \(\frac{1}{sin^2θ}\)

cot²θ - \(\frac{1}{sin^2θ}\) = cot²θ - cosec²θ 

Also, we know that 1 + cot² θ = cosec²θ

⇒ cot²θ – cosec²θ = – 1

cot²θ - \(\frac{1}{sin^2θ}\) = cot²θ – cosec²θ = – 1

(1 + cot²θ) sin²θ = 1

L.H.S. = (1 + cot²θ) × sin² θ

= (cosec² θ) × sin² θ

(Using identity 1 + cot² θ = cosec² θ)

= 1/ sin²θ × sin² θ

= 1

= R.H.S.

Hence Proved.

(sec²θ − 1) (cosec²θ − 1) = 1

L.H.S. = (sec² θ – 1)(cosec² θ – 1)

= (tan²θ) × cot²θ

(using identity 1 + cot² θ = cosec² θ and 1 + tan² θ = sec² θ)

= tan²θ x 1/tan²θ

= 1

= R.H.S.

Hence Proved.

(sec²θ − 1) cot²θ = 1

L.H.S. = (sec² θ – 1) × cot² θ

= (tan²θ) x cot²θ

(using identity 1 + tan² θ = sec² θ)

= 1/cot²θ x cot²θ

= 1

= R.H.S.

Hence Proved.

(1 + cos θ) (1 − cos θ) (1 + cot²θ) = 1

LHS: (1 + cos θ) (1 − cos θ) (1 + cot²θ)

= (1 – cos² θ) × cosec² θ

(Using sin² θ + cos² θ = 1)

= (sin² θ) × cosec² θ

= sin² θ x 1/sin² θ

= 1

= R.H.S.

Hence Proved

Cos 1 cos 2 … cos 180

= cos 1 cos 2 … cos 90… cos 180

= cos 1 cos 2 … 0 … cos 180

= 0 

Answer is (D) -cosec θ

Using identity, 

sin² θ + cos² θ = 1  ⇒ sin² θ = 1 – cos² θ 

Taking L.H.S, 

L.H.S = cosec θ √(1 – cos² θ) 

= cosec θ √( sin² θ) 

= cosec θ x sin θ 

= 1 

= R.H.S 

Hence Proved

x = acos³θ ⇒ \(\frac{x}{a}\) = cos³θ 

y = bsin³θ ⇒ \(\frac{y}{b}\) = sin³θ

Now, \(\Big(\frac{x}{a}\Big)^{2/3}+\Big(\frac{y}{b}\Big)^{2/3}\) = (cos³θ)^2/3 + (sin³θ)^2/3

= cos²θ +sin²θ 

= 1

\(\Big(\frac{x}{a}\Big)^{2/3}+\Big(\frac{y}{b}\Big)^{2/3}\) = 1

 cosecθ\( \sqrt{1-cos^2θ} \) =  \(\frac{1}{sinθ}\sqrt{sin^2θ}\)

= \(\frac{1}{sinθ}\times sinθ\)

= 1

Hence Proved.

(sec²θ -1((cosec²θ-1)=tan²θ x cot²θ

= 1

Hence Proved.

\(\frac{1+sinθ}{1-sinθ}\) = \(\frac{1+sinθ}{1-sinθ}\times \frac{1+sinθ}{1+sinθ}\) 

= \(\frac{(1+sinθ)^2}{1-sin^2θ}\) 

= \(\frac{(1+sinθ)^2}{cos^2θ}\) 

= sec²θ + tan²θ

=  tan²θ + 1 +  tan²θ

= 2 tan²θ + 1

= 2 x \(\Big(\frac{12}{5}\Big)^2+1\)   \(\Big(∵tanθ=\frac{12}{5}\Big)\)

= \(\frac{288}{25}+1\)

= \(\frac{288+25}{25}\) = \(\frac{313}{25}\)

By using the identity, 

cosec² A – cot² A = 1 

⇒ cosec² A = cot² A + 1 

Taking, L.H.S = (1 + cot² A) sin² A 

= cosec² A sin² A

= (cosec A sin A)² 

= ((1/sin A) × sin A)² 

= (1)² 

= 1 

= R.H.S 

Hence Proved

Taking the L.H.S, 

(1 – cos² A) cosec² A 

= (sin² A) cosec² A [∵ sin² A + cos² A = 1 ⇒1 – sin² A = cos² A] 

= 1² 

= 1 

= R.H.S 

Hence Proved

An equation that is true for all values of the variables involved is said to be an identity. 

For example: 

a² – b² = (a – b) (a + b) 

sin²θ + cos²θ = 1

From trig. Identities we have, 

sec² θ − tan² θ = 1 

On cubing both sides, 

(sec²θ − tan²θ)³ = 1

sec⁶ θ − tan⁶ θ − 3sec² θ tan² θ(sec² θ − tan² θ)= 1 

[Since, (a – b)³ = a³ – b³ – 3ab(a – b)] 

sec⁶ θ − tan⁶ θ − 3sec² θ tan² θ = 1 

⇒ sec⁶ θ = tan⁶ θ + 3sec² θ tan² θ + 1 

Hence, L.H.S = R.H.S 

Hence proved

Taking L.H.S 

= (cosec θ + sin θ)(cosec θ – sin θ) 

On multiplying we get, 

= cosec² θ – sin² θ 

= (1 + cot² θ) – (1 – cos² θ) [Using cosec² θ − cot² θ = 1 and sin² θ + cos² θ = 1] 

= 1 + cot² θ – 1 + cos² θ 

= cot² θ + cos² θ 

= R.H.S 

Hence Proved

To find: 9 sec²A – 9 tan²A 

Consider 9 sec²A – 9 tan²A = 9 (sec²A – tan²A) 

∵ 1 + tan²A = sec²A 

∴ 9 sec²A – 9 tan²A = 9 (sec²A – tan²A) 

= 9 (1 + tan²A – tan²A) = 9

Given: x a sec θ cos ϕ 

Squaring both sides, we get 

x² = a²sec²θ cos²ϕ 

and y = b sec θ sin ϕ

Squaring both sides, we get 

y² = b² sec²θ sin²ϕ 

And z = c tan θ 

⇒ z² = c²tan²θ

⇒ tan²θ = \(\frac{z^2}{c^2}\)  ........(i)

To find: \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) 

Consider \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)  = \(\frac{a^2sec^2θcos^2Φ}{a^2} + \frac{b^2sec^2θsin^2Φ}{b^2}\)

= sec²θ cos²ϕ + sec²θ sin²ϕ 

= sec²θ (cos²ϕ + sin²ϕ) 

= sec²θ [∵ sin²ϕ + cos²ϕ = 1] 

= 1 + tan²θ [∵ 1 + tan²θ = sec²θ]

= \(1+\frac{z^2}{c^2}\) 

\(\frac{1-sinθ}{1+sinθ}\) = \(\frac{1-sinθ}{1+sinθ}\times\frac{1-sinθ}{1-sinθ}\)

=   \(\frac{(1-sinθ)^2}{1-sin^2θ}\)  

=    \(\frac{(1-sinθ)^2}{cos^2θ}\)​​

= ​​​​​ \(\Big(\frac{1-sinθ}{cosθ}\Big)^2\)  

= \(\Big(\frac{1}{cosθ}-\frac{sinθ}{cosθ}\Big)^2\)

= (secθ - tanθ)²

Hence Proved.

  sin²A + \(\frac{1}{1+tan^2A}=sin^2A+\frac{1}{sec^2 A}\)

= sin² A + cos²A

= 1

Hence Proved.