Introduction to Trigonometry

NCERT Exercise 8.3 Solution part1

Trigonometric Ratios of Complementary Angles

sin (90^o – A) = cos A

cos (90^o – A) = sin A

tan (90^o – A) = cot A

cot (90^o – A) = tan A

sec (90^o – A) = cosec A

Solution of Questions of NCERT Exercise 8.3 part:1

Question (1) Evaluate:

(i) ^{sin 18o}/_{cos 72^o}

Solution:

We know that, sinA = cos (90^o – A)

∴ sin 18^o = cos (90^o – 18^o)

⇒ sin 18^o= cos 72^o

∴ ^{sin 18o}/_{cos 72^o}

After substituting the sin 18^o= cos 72^o

= ^{cos 72o}/_{cos 72^o}

= 1 Answer

(ii) ^{tan 26o}/_{cot 64^o}

Solution:

We know that, cot A = tan (90^o – A),

∴ cot 64^o = tan (90^o – 64^o)

⇒ cot 64^o = tan 26^o

Given,

^{tan 26o}/_{cot 64^o}

Thus, After substituting cot 64^o = tan 26^o, we get

= ^{tan 26o}/_{tan 26^o}

= 1 Answer

(iii) cos 48^o – sin 42^o

Solution:

We know that

sin A = cos (90 – A)

∴ sin 42^o = cos (90^o – 48^o)

⇒ sin 42^o = cos 48^o

Here, given in question,

cos 48^o – sin 42^o

Thus, after replacing sin 42^o = cos 48^o, we get

= cos 48^o – cos 48^o

= 0 Answer

(iv) cosec 31^o – sec 59^o

Solution:

We know that, sec A = cosec (90^o – A)

∴ sec 59^o = cosec (90^o – 59^o)

⇒ sec 59^o = cosec 31^o

Here given in the question,

cosec 31^o – sec 59^o

Thus, after substituting the value of sec 59^o = cosec 31^o, we get

= cosec 31^o – cosec 31^o

= 0 Answer

Question (2) Show that,

(i) tan 48^o tan 23^o tan 42^o tan 67^o = 1

Solution:

We know that,

tan A = cot (90^o – A)

∴ tan 48^o = cot (90^o– 48^o)

⇒ tan 48^o = cot 42^o - - - - - (i)

And, tan 67^o = cot (90^o – 67^o)

⇒ tan 67^o = cot 23^o - - - - - (ii)

Given in question,

LHS = tan 48^o tan 23^o tan 42^o tan 67^o

After substituting values of tan 48^o and tan 67^o from equation (i) and (ii), we get

= cot 42^o . tan 23^o . tan 42^o . cot 23^o

= ¹/_{tan 42^o} × tan 23^o × tan 42^o × ¹/_{tan 23^o}

= ^{tan 23o}/_{tan 42^o} × ^{tan 42o}/_{tan 23^o}

= 1 Proved

(ii) cos 38^o cos 52^o – sin 38^o sin 52^o = 0

Solution:

We know that,

cos A = sin (90^o – A)

∴ cos 38^o = sin (90^o – 38^o)

⇒ cos 38^o = sin 52^o - - - - - - (i)

And, cos 52^o = sin (90^o – 52^o)

⇒ cos 52^o = sin 38^o - - - - - - (ii)

Now, LHS in question,

cos 38^o cos 52^o – sin 38^o sin 52^o

After substituting values of cos 38^o and cos 52^o from equation (i) and (ii), we get

= sin 52^o sin 38^o – sin 38^o sin 52^o

= 0 Answer

MCQs Test

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