Trigonometric Identities Class 10 Maths Previous Year Questions (Most Repeated) | CBSE Boards 2023

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Welcome to the new episode of our new series "PYQs".
In this session, Khushbu ma'am will teach you the 10 most repeated previous years questions from on Trigonometric Identities to make you exam ready.

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Trigonometric Identities

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[BYJUSClass]

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[prabhasharma]

n(m²-1) given n=secA+cosecA and m=sinA+cosA
Let's put the values
(secA+cosecA)[(sinA+cosA)²-1]

Taking LCM in the first bracket and put sin²A+cos²A=1 in the second bracket

sinA+cosA/cosAsinA ×2sinAcosA. {1and-1cancelled}
2(sinA+cosA). {cosAsinA cancelled from 2cosAsinA}
Since sinA+cosA=m
Therefore we end up with 2m
Hence proved .
Answered by PRIYANSHU SHARMA
Session was

[naveenbunny]

n(m^2-1) =2m
LHS=n(m^2-1)
From the question we get
sec(theta) +cosec(theta){[sin(theta) +cos(theta)]^2-1}
=sec(theta) +cosec(theta) [sin^2(theta) +cos^2(theta) +2sin(theta)cos(theta) -(sin^2(theta)+cos^2(theta) ]
]
=1/cos(theta) +1/sin([2sin(theta). cos(theta)]
=sin(theta)+cos(theta)
=2[sin(theta)+cos(theta)]
=2m
Therefore LHS=RHS
Hence proved

[sanjayscience]

चीते की चाल
बाज की उड़ान
और Class 10th के students की त्यारी पर कोई शक नही करते l I hope everyone get 100/100 in mathematics. Thank 🌹 you so much provide this lectures

[priyankatiwari]

Ma'am please upload important question from quadratic and AP ....really need those sessions ...

[yashmhoparkar]

Hw question's answer =n (m²-1)
(SecA +cosecA)((sinA+cosA)²-1)
(SecA+cosecA)(2sinA.cosA)
(1/cosA+1/sinA)(2sinA.cosA)
2sinA.cosA/cosA +2sinA.cosA/sinA
=2sinA+2cosA
2(sinA+cosA)
I.e 2m
Hence proved🙂🙂🙂🙂🙂🙂🙂
Thanks mam and byjus team for such amazing session

[atmadeeproy]

1:14:36 Yes ma'am I have got the answer
First I converted the value of 'n' in terms of cos and sin from where I got the value of (sin theta * cos theta), then I proceeded with LHS part of proof where I replaced the value of (sin theta * cos theta) and finally got the answer
Really helpful session 👍🏻❤️

[anilgovind]

Thanku ma'am,
The same questions came in my pre model, and i for 18 out of 20
Homework,
n(m²+1)=
✓n=sin+cos
✓m=sec+cosec
Hence
(Sec +cosec)((Sin+cos)²-1)=

(Sec+cos)(1+2sincos-1)
(2sincos)/(¹/cos+¹/sin)
(2sincos)((sin+cos)/sincos)
2((cos+sin))
Add we know ✓m=cos +sin
n(m²-1)=2m
Hence proved ✓✓

Thank you ma'am for the great session ☺️💫

[naseera]

Homework
Sec theta + cosec theta ((sin theta+cos theta)^2)-1
Sec theta + cosec theta (sin^2+cos^2+2sinocoso - 1)
*Use a+b whole sq identity*
Sec theta+cosec theta (1+2sinocoso-1)
1/cos theta + 1/sin theta (2 sinocoso)
Sin theta + cos theta/sinocoso ( 2 sinocoso)

2sin theta + 2 cos theta
>>> 2m >>>RHS

[inayasharmavolgs]

Hey byjus!!
Loved the session thanks for your efforts and help.
*Please bring a session for word problems of arithmetic progression*.

[sunitabora]

No mam 2 indentity are deleted
1+tan²A=sec²A or
cot²A+1=cosec²A
Only this indentity is given 👇
sin²A+cos²A=1

[priyanshusinghrajput]

Mam Homework question was so easy one, Only we have to do is that to put the given data into the question and then by using identity and maths calculation it will cane into 6-7 steps... Thanks For The SESSION MA"AM.

[mlrashah]

Yes Ma'am, I was able to show that n(m^2 -1)=2m. The identities used in this question are (a+b)^2 and sin^2theta+cos^2theta=1 along with taking sectheta and cosectheta in terms of their reciprocals costheta and sintheta respectively. Also, using the value of m provided .

[logogameryt]

n(m²-1)=2m
Replace all the values of n and m we get

LHS=
SecA+cosecA{(sinA+cosA)²-1}

=1/cosA +1/sinA {(sin²A+cose²A+2sinAcosA)-1}


=SinA+cosA(1+2-1)
=SinA+cosA(2)
=2(sinA+cosA)
Mam I have replaced theta to A

[satishsathya]

n(m²-1)=2m
LHS = n(m²-1)
= Sec theta +cosec theta {(sin theta +cos theta)²-1}
= 1/cos theta + 1/sin theta(sin ²theta + cos² theta+2sin theta cos theta -1)
= Sin theta+cos theta/sin theta cos theta ( 1 +2 sin theta cos theta-1)
=Sin theta cos theta/sin theta cos theta(2 sin theta cos theta)
= Sin theta +cos theta× 2
= m. × 2
= 2m
= RHS 👍👍
Amazing session ma'am

[subhasritapadhy]

Mam you are teaching is really fantastic

[medicosis_untannable]

{Fist we will put values of m and n in the 1st eq}
sec+cosec[(sin+cos)^2-1]= 2(sin+cos)

Solving LHS
{Then we will apply identity of sin^2+cos^2=1}
{Sec and cosec in terms of cos and sin respectively and then LCM}

[(Sin+cos)/sin cos]× 2sin cos
{Cancel out sin cos with 2 sin cos}

And therefore, 2[sin + cos] = 2[sin + cos]
LHS = RHS

[lavanyakt]

42:52 mam we cant solve the debominator as such cuz
1=sec^2- tan^2 ==(secA+tanA)(secA-tanA)
Hence we cant take common because denominator we have tanA-SecA not secA-tanA

[ruksharkhan]

1:03:29 mam why if cos square thita + sin sqaure thita should be 1 why you didn't used it pls reply

[tanveer]

Ma'am literally the session was so amazing I am thankful towards Byjus. 😇😇😇