CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

LHS =(sin A + sec A)² + (cos A + cosec A)²
{tex}= \left( \sin A + \frac { 1 } { \cos A } \right) ^ { 2 } + \left( \cos A + \frac { 1 } { \sin A } \right) ^ { 2 }{/tex}
{tex}= \sin ^ { 2 } A + \frac { 1 } { \cos ^ { 2 } A } + 2 \frac { \sin A } { \cos A } + \cos ^ { 2 } A{/tex}{tex}+ \frac { 1 } { \sin ^ { 2 } A } + 2 \frac { \cos A } { \sin A }{/tex}
= {tex}sin^2A+cos^2A{/tex} + {tex}\frac { 1 } { \sin ^ { 2 } A } + \frac { 1 } { \cos ^ { 2 } A }{/tex}{tex}+ 2 \left( \frac { \sin A } { \cos A } + \frac { \cos A } { \sin A } \right){/tex}
= 1 + {tex}\frac { \sin ^ { 2 } A + \cos ^ { 2 } A } { \sin ^ { 2 } A \cos ^ { 2 } A } + 2 \left( \frac { \sin ^ { 2 } A + \cos ^ { 2 } A } { \sin A \cos A } \right){/tex}
= 1 + {tex}\frac { 1 } { \sin ^ { 2 } A \cos ^ { 2 } A } + \frac { 2 } { \sin A \cos A }{/tex}
= {tex}\left( 1 + \frac { 1 } { \sin A \cos A } \right) ^ { 2 }{/tex}
= {tex}(1+secAcosecA)^2{/tex}

^=RHS

0Thank You

ANSWER

Ghhydga

Posted by Aaditya Prabal Chawla 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Ncert. Solution of. All chapter answer key

Posted by Vishal Singh 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Check NCERT solutions here : <a href="https://mycbseguide.com/ncert-solutions.html">https://mycbseguide.com/ncert-solutions.html</a>

0Thank You

ANSWER

Find the product zero of the polynomial p(x)=ax^3+bx^2+cx+d

Posted by Chinmoy Das 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Chinmoy Das 7 years, 5 months ago

0Thank You

ANSWER

A man arranges to pay off debt of rs3600 by 40 annual installments which form an arthmetic series .when 30 installments are paid he dies leaving one third of the debt upaid what is the value of firt installment

Posted by Kunj Khemka 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Let first instalment(a) = Rs a
Let common difference = Rs d
Given,
Amount of 40 instalments = Rs 3600
{tex}\Rightarrow \frac { 40 } { 2 } [ 2 a + ( 40 - 1 ) d ] = 3600{/tex}
{tex}\Rightarrow{/tex}20[2a + 39d] = 3600
{tex}\Rightarrow{/tex}2a + 39d = 180..............(i)
and, Amount of 30 instalments = Rs 2400
{tex}\Rightarrow \frac { 30 } { 2 } [ 2 a + ( 30 - 1 ) d ] = 2400{/tex}
{tex}\Rightarrow{/tex}15[2a + 29d] = 2400
{tex}\Rightarrow{/tex}2a + 29d = 160..............(ii)
Subtracting (ii) from (i),
2a + 39d - 2a - 29d = 180 - 160
{tex}\Rightarrow{/tex} 10d = 20
{tex}\Rightarrow \quad d= \frac { 20 } { 10 } = 2{/tex}
Putting value of d in eq(i),
2a + 39{tex}\times{/tex} 2 = 180
{tex}\Rightarrow{/tex}2a + 78 = 180
{tex}\Rightarrow{/tex}2a = 180 - 78
{tex}\Rightarrow{/tex}2a = 102
{tex}\Rightarrow a = \frac { 102 } { 2 } = 51{/tex}
Therefore, Value of first instalment = Rs 51

0Thank You

ANSWER

Prove that tan theta + sec theta minus 1 over over 10 theta minus sec theta + 1 is equal to cos theta over 1 minus sin theta

Posted by Arun Chauhan Taniya 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

According to the question,

L.H.S. = {tex}\frac { \sec \theta + \tan \theta - 1 } { \tan \theta - \sec \theta + 1 }{/tex}
{tex}= \frac { ( \sec \theta + \tan \theta ) - \left( \sec ^ { 2 } \theta - \tan ^ { 2 } \theta \right) } { ( \tan \theta - \sec \theta + 1 ) }{/tex} [{tex}\because{/tex} sec²{tex}\theta{/tex} - tan²{tex}\theta{/tex} = 1]
{tex}= \frac { ( \sec \theta + \tan \theta ) [ 1 - ( \sec \theta - \tan \theta ) ] } { ( \tan \theta - \sec \theta + 1 ) }{/tex}
{tex}= \frac { ( \sec \theta + \tan \theta ) ( \tan \theta - \sec \theta + 1 ) } { ( \tan \theta - \sec \theta + 1 ) } = ( \sec \theta + \tan \theta ){/tex}
{tex}= \left( \frac { 1 } { \cos \theta } + \frac { \sin \theta } { \cos \theta } \right) = \frac { ( 1 + \sin \theta ) } { \cos \theta } = \frac { ( 1 + \sin \theta ) } { \cos \theta } \times \frac { ( 1 - \sin \theta ) } { ( 1 - \sin \theta ) }{/tex}
{tex}= \frac { \left( 1 - \sin ^ { 2 } \theta \right) } { \cos \theta ( 1 - \sin \theta ) } = \frac { \cos ^ { 2 } \theta } { \cos \theta ( 1 - \sin \theta ) } = \frac { \cos \theta } { ( 1 - \sin \theta ) }{/tex} = R.H.S.
{tex}\therefore{/tex} {tex}\frac { \sec \theta + \tan \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { \cos \theta } { ( 1 - \sin \theta ) }{/tex}

0Thank You

ANSWER

How can you prove theorm

Posted by Amit Kumar Rawat 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Hardik X Animack 7 years, 5 months ago

which theorem

1Thank You

ANSWER

Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Posted by Jamin Agarawal 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Chinmoy Das 7 years, 5 months ago

Let the no. of Rs.50 notes= x Let the no. of Rs.100 note= y A/q x+y=25.........(1) 50x+100y=2000 x+2y=40 (dividing all by 50)...........(2) Now, (1) - (2) By elimination method x+y=25 x+2y=40 .................. -y=-15 => y = 15 Putting y=15 in (1) x+15=25 => x = 25 - 15 => x = 10 Therefore No. of Rs.50 notes =10 No. of Rs.100 notes = 15

1Thank You

ANSWER

Sir i really want to pass my exam i have hardly 10 days left i dont know anything sir please help please

Posted by Robin Singh 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Rahul Yadav 7 years, 5 months ago

Concentration power rakho And think i can do it And it is too eazy

7Thank You

ANSWER

A rectangular garden 10m by 16m is to be surrounded by a concrete wall of uniform width. If Area of the wall is 120 m²,find the width of the wall.

Posted by Sarvesh Kumar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

2 m

0Thank You

ANSWER

By splitting the middle term what do we get in this question : 4x raise to 2 - 12x + 9

Posted by Viraj Rathod 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Rahul Yadav 7 years, 5 months ago

4x^2-6x-6x+9 2X(2x+3) -3(2x+3) (2x+3)(2x-3) it is zero when 2x+3=0 & 2x-3=0 X=-3/2 & X=3/2

4Thank You

ANSWER

Posted by Murari Verma 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Find the value of (-1)+(-1)2n+(-1)2n+1+(-1)4n-1where n is postive odd integer

Posted by Ankit Ranjan 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Ankit Ranjan 7 years, 5 months ago

Answer

0Thank You

ANSWER

If the mean of following frequency distrubuition is 91.find the missing frequencies f1 and f2 Classes 0-30 30-60 60-90 90-120 120-150 150-180 total Frequency 12 21 f1 52 f2 11 150

Posted by Manmeet Singh 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

example 8.4 5th sum

Posted by Palak Jain 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

What is cos67 - sin 23 ans

Posted by Shyam Janam 7 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Ashray H. 7 years, 5 months ago

cos67-sin23 cos67 - cos(90 - 23) = cos67 - cos67 = 0

3Thank You

Divyansh Dubey Ji 7 years, 5 months ago

Thanks

1Thank You

ANSWER

What is the prime factor of 146

Posted by Yashvi Patel 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Elsa B 7 years, 5 months ago

146 is a composite number. Factor pairs: 146 = 1 x 146 or 2 x 73. Factors of 146: 1, 2, 73. Prime factorization: 146 = 2 x 73.

2Thank You

ANSWER

For what value of p will the following pair of linear equations have infinitely many solutions? (p-q)x+3y=p;px+py=12

Posted by Vibu Guna 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

How to solve completing method

Posted by Harsh Vishwakarma 7 years, 5 months ago

CBSE > Class 10 > Mathematics

3 answers

Aditya Kumae 7 years, 5 months ago

Step 4 : now for the value of x

0Thank You

Aditya Kumae 7 years, 5 months ago

Step3: add (1/2×coefficient of x) to both RHS and LHS

0Thank You

Aditya Kumae 7 years, 5 months ago

Step1: arrange them in their standard form Step2: coefficient of x² should be 1(if any other then divide by its coefficient to all term s of equation

0Thank You

ANSWER

Find the smallest number which when increase by 17 excatly divisible by both 520 and 468

Posted by Shrutika Magdum 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

The smallest number divisible by 520 and 468 = LCM(520,468)
Prime factors of 520 and 468 are :
520 = 2³ {tex}\times{/tex} 5 {tex}\times{/tex} 13
468 = 2 {tex}\times{/tex} 2 {tex}\times{/tex} 3 {tex}\times{/tex} 3 {tex}\times{/tex} 13
Hence LCM(520,468) = 2³ {tex}\times{/tex} 3²{tex}\times{/tex} 5 {tex}\times{/tex} 13 = 8 {tex}\times{/tex} 9{tex}\times{/tex}5{tex}\times{/tex}13 = 4680
Now the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
=LCM(520,468)-17
=4680-17
=4663