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Ask QuestionPosted by Shubhankar Mishra 3 years, 11 months ago
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Posted by Zeeshan Shaikh 3 years, 10 months ago
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Shubhankar Mishra 3 years, 11 months ago
Ajad Kumar 3 years, 11 months ago
Posted by Mrs Queen 3 years, 11 months ago
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Shubhankar Mishra 3 years, 11 months ago
Posted by Sukhveer Singh 4 years ago
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Posted by Manohar Patel 4 years ago
- 1 answers
Yogita Ingle 4 years ago
tan10°tan15°tan 75° tan80°
= tan10 tan15 tan(90-75) tan (90-80)
= tan10 tan15 cot 15 cot10
as we know tan theta * cot theta = 1
so
tan10 cot10 tan15 cot 15
= 1
Posted by Sanskar Kanojiya 3 years, 11 months ago
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Posted by Salvitripathi73@Gmail.Com Tripathi 4 years, 1 month ago
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Posted by Ajit Ajs Ajit Ajs 4 years, 2 months ago
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Posted by Aditya Vishwakarma 4 years, 2 months ago
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Posted by Salma Shah 4 years, 2 months ago
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Yogita Ingle 4 years, 2 months ago
Real numbers can be defined as the union of both the rational and irrational numbers. They can be both positive or negative and are denoted by the symbol “R”. All the natural numbers, decimals and fractions come under this category. See the figure, given below, which shows the classification of real numerals.
Posted by Abhishek Mahto 4 years, 3 months ago
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Posted by Zia Yd 4 years, 3 months ago
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Posted by Kabyashree Bordoloi 4 years, 4 months ago
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Posted by Damini Suman 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
x + y = 5
x = 5 - y......... (i)
2x - 3y = 4 .......... (ii)
Put (i) in (ii) , we get
2(5 - y) -3y = 4
10 - 2y - 3y = 4
10 - 5y = 4
-5y = 4 - 10
-5y = -6
y = 6/5
Put y = 6/5 in (i) we get
x = 5 - 6/5
x = 19/5
Posted by Anushka Somvanshi 4 years, 4 months ago
- 1 answers
Gaurav Seth 4 years, 4 months ago
We know that,
Dividend = Divisor × Quotient + Remainder
⇒ Dividend – Remainder = Divisor × Quotient
⇒ Dividend – Remainder is always divisible by the divisor.
Now, it is given that f(x) when divided by x2 – 2x + k leaves (x + a) as remainder.
So, for f(x) to be completely divisible by x2 – 2x + k, remainder must be equal to zero
⇒ (–10 + 2k)x + (10 – a – 8k + k2) = 0
⇒ –10 + 2k = 0 and 10 – a – 8k + k2 = 0
⇒ k = 5 and 10 – a – 8 (5) + 52 = 0
⇒ k = 5 and – a – 5 = 0
⇒ k = 5 and a = –5
Posted by Ashish Gupta 4 years, 4 months ago
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Posted by Yuvraj Yadav 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
(cos 0+sin 45+sin 30)(sin 90-cos 45+cos 60)
=
=
=
/* By algebraic identity
(a-b)(a+b) = a²-b² */
=
=
=
Posted by Thedev Mishra 4 years, 4 months ago
- 0 answers
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